What is an irrational algebraic function?

An irrational algebraic function is a function that involves algebraic expressions with roots or fractional exponents. These functions cannot be expressed as a ratio of polynomials. Examples include √x, ³√(x²+1), or x^(1/4).

Why is integrating irrational algebraic functions often challenging?

Integrating irrational algebraic functions is often challenging because they don't follow standard integration rules for polynomial or rational functions. They often require special techniques like substitution or trigonometric substitutions to simplify the integrand before integration.

What is the first step in integrating an irrational algebraic function?

The first step is to identify the type of irrationality present in the function. This could be a square root, cube root, or a more complex fractional exponent. Understanding the nature of the irrationality helps in choosing the appropriate integration technique.

How does substitution help in integrating irrational algebraic functions?

Substitution helps by transforming the irrational function into a simpler, often rational function. By carefully choosing a substitution, we can eliminate the irrationality and make the integration more straightforward. After integration, we then substitute back to get the final result in terms of the original variable.

What is the significance of the degree of the radicand in integration?

The degree of the radicand (the expression under the root) is crucial in determining the integration method. If the degree of the radicand is even and the root index is odd, trigonometric substitution is often effective. For other cases, different substitution methods may be more appropriate.

How do you integrate √(ax + b)?

To integrate √(ax + b), we can use the substitution u = √(ax + b). This transforms the integral into a simpler form. The steps are:

What is the role of trigonometric substitutions in integrating irrational functions?

Trigonometric substitutions are useful for integrating certain forms of irrational functions, particularly those involving square roots of quadratic expressions. They work by converting the irrational expression into trigonometric functions, which often simplify the integration process. After integration, we convert back to the original variable.

When should you use the substitution x = a sin θ in integration?

The substitution x = a sin θ is typically used when the integrand contains √(a² - x²). This substitution transforms the square root into a cosine function, which is often easier to integrate. It's particularly useful for integrals involving circles or ellipses.

How do you integrate expressions containing ³√x?

To integrate expressions containing ³√x, a common approach is to use the substitution u = ³√x. This transforms the cube root into a simpler polynomial in u. The steps are:

What is the strategy for integrating (x² + a²)^(1/4)?

For integrating (x² + a²)^(1/4), a useful strategy is to use the substitution u = (x² + a²)^(1/4). This transforms the fourth root into a linear term in u. The process involves:

How does the method of integration differ for even and odd roots?

The method of integration often differs for even and odd roots due to their different properties:

What is the significance of the discriminant in integrating irrational functions?

The discriminant of a quadratic expression ax² + bx + c (which is b² - 4ac) plays a crucial role in determining the appropriate substitution for integrating irrational functions involving square roots. The sign of the discriminant (positive, negative, or zero) guides the choice between trigonometric, hyperbolic, or rational substitutions.

How do you integrate √(x² + a²)?

To integrate √(x² + a²), we typically use the trigonometric substitution x = a tan θ. The steps are:

What is the difference between integrating √(x² - a²) and √(a² - x²)?

The integration of √(x² - a²) and √(a² - x²) requires different approaches:

How do you handle integrals involving both rational and irrational parts?

For integrals involving both rational and irrational parts, a common strategy is to:

What is the role of partial fractions in integrating irrational functions?

Partial fractions decomposition can be useful after making a substitution that transforms an irrational function into a rational one. Once the irrational function is converted to a rational function through substitution, partial fractions can be used to break down complex rational expressions into simpler terms that are easier to integrate.

How do you integrate expressions of the form (ax + b)^(n/m) where n and m are integers?

To integrate expressions of the form (ax + b)^(n/m):

What is the importance of recognizing standard forms in irrational integrals?

Recognizing standard forms in irrational integrals is crucial because:

How do you integrate √(1 + x²)?

To integrate √(1 + x²), we use the substitution x = tan θ. The steps are:

What is the significance of the substitution u = x + √(x² + a²) in integration?

The substitution u = x + √(x² + a²) is significant because:

How do you integrate expressions involving fourth roots, like ⁴√(x²+1)?

To integrate expressions with fourth roots like ⁴√(x²+1):

What is the role of hyperbolic functions in integrating certain irrational expressions?

Hyperbolic functions are useful in integrating certain irrational expressions because:

How do you approach integrating nested radicals like √(x + √x)?

To integrate nested radicals like √(x + √x):

What is the strategy for integrating irrational functions with polynomial numerators?

For integrating irrational functions with polynomial numerators:

Integration of Irrational Algebraic Function

Maths
Integrals
Integration of Irrational Algebraic Function

Integration of Irrational Algebraic Function

Komal MiglaniUpdated on 02 Jul 2025, 08:09 PM IST

Integration of Irrational Algebraic Function is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These integration concepts have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

In this article, we will cover the concept of Integration of Irrational Algebraic Functions. This concept falls under the broader category of Calculus, a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), four questions have been asked on this concept, including one in 2020, one in 2021, and two in 2022.

Integration of irrational functions

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given. The rate of change of a quantity y concerning another quantity x is called the derivative or differential coefficient of y concerning x. Geometrically, the Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point.

Irrational Functions

An irrational equation is an equation where the variable is inside the radical or the variable is a base of power with fractional exponents. A function with one or more irrational equaitons are called irrational functions.

(a) Integrals of the Form
(i) $\int \frac{1}{(a x+b) \sqrt{p x+q}} d x$
(ii) $\int \frac{a x+b}{\sqrt{p x+q}} d x$
(iii) $\int \frac{\sqrt{p x+q}}{a x+b} d x$
(iv) $\int \frac{1}{\left(a x^2+b x+c\right) \sqrt{p x+q}} d x$

To evaluate this type of integrals, put $p x+q=t^2$

Evaluate: $\int \frac{d x}{(x+1) \sqrt{x+2}}$
Let $\mathrm{I}=\int \frac{d x}{(x+1) \sqrt{x+2}}$
Substitute: $x+2=t^2 \quad \Rightarrow \quad d x=2 t d t$

$
\begin{aligned}
& \Rightarrow \int \frac{d x}{x+1(\sqrt{x+2})}=\int \frac{2 t d t}{\left(t^2-1\right) \sqrt{t^2}}=2 \int \frac{d t}{t^2-1} \\
& =\frac{2}{2} \log \left|\frac{t-1}{t+1}\right|+C=\log \left|\frac{\sqrt{x+2}-1}{\sqrt{x+2+1}}\right|+C
\end{aligned}
$

(b) Integrals of the Form
(i) $\int \frac{1}{(p x+q) \sqrt{a x^2+b x+c}} d x$

To evaluate this type of integral, put $p x+q=1 /$ t

(c) Integrals of the Form
(i) $\int \frac{1}{\left(\mathrm{ax}^2+\mathrm{b}\right) \sqrt{\mathrm{px}^2+\mathrm{q}}} d \mathrm{x}$

To evaluate this type of integral, put $x=\frac{1}{t}$

Solved Examples Based On Integration Of Irrational Functions

Example 1: Evaluate $\int \frac{d x}{(x-1) \sqrt{x^2+x+1}}, \mathrm{x}>1$.

$\begin{aligned}
& -\frac{1}{\sqrt{3}} \ln \left|\frac{1}{x-1}-\frac{1}{2}+\sqrt{\frac{12\left(\frac{1}{x-1}+\frac{1}{2}\right)^2+1}{12}}\right|+c \\
& \text { 2) } \frac{1}{\sqrt{3}} \ln \left|\frac{1}{x-1}-\frac{1}{2}+\sqrt{\frac{12\left(\frac{1}{x-1}+\frac{1}{2}\right)^2+1}{12}}\right|+c \\
& \frac{1}{\sqrt{3}} \ln \left|\frac{1}{x-1}+\frac{1}{2}+\sqrt{\frac{12\left(\frac{1}{x-1}+\frac{1}{2}\right)^2+1}{12}}\right|+c \\
& \text { 3) } \\
& \text { 4) }-\frac{1}{\sqrt{3}} \ln \left|\frac{1}{x-1}+\frac{1}{2}+\sqrt{\frac{12\left(\frac{1}{x-1}+\frac{1}{2}\right)^2+1}{12}}\right|+c
\end{aligned}$
Solution

As we learnt

The integration in the form

$\int_{\text {(i) }} \frac{d x}{(p x+q) \sqrt{a x^2+b x+c}}$

(ii) $\int \frac{d x}{(p x+q) \sqrt{a x+b}}$
$\int_{\text {(iii) }} \frac{(a+b x)^m}{(p+q x)^n} d x$
- wherein

Working rule.
$(\mathrm{i}) \rightarrow_{\text {put }}(p x+q)=\frac{1}{t}$
(ii) $\rightarrow$ put $(a x+b)=t^2$
(iii) $\rightarrow$ put $(p+q x)=t$

Put $\mathrm{x}-1=1 / \mathrm{t}$ and $d x=-1 / t^2 \mathrm{dt}$.
Invalid Equation

$=-\frac{1}{\sqrt{3}} \ln \left|(t+1 / 2)+\sqrt{\left(t+\frac{1}{2}\right)^2+\frac{1}{12}}\right|+c$

Invalid Equation

4) None of these

Example 2: $\int \frac{x}{(x-3) \sqrt{x+1}} d x$
1) $-2 \sqrt{x+1}-\frac{3}{2} \ln \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+c$
2) $2 \sqrt{x+1}-\frac{3}{2} \ln \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+c$
3) $2 \sqrt{x+1}+\frac{3}{2} \ln \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+c$Solution

Solution

As we learned

The working rule is

put $(a x+b)=t^2$
Put $x+1=t^2$. We get

$\begin{aligned}
& \int \frac{x}{(x-3) \sqrt{x+1}} d x \\
& \int \frac{x-3+3}{(x-3) \sqrt{x+1}} d x=\int \frac{x-3}{(x-3) \sqrt{x+1}} d x+3 \int \frac{1}{(x-3) \sqrt{x+1}} d x \\
& \int \frac{1}{\sqrt{x+1}} d x+3 \int \frac{1}{(x-3) \sqrt{x+1}} d x \\
& \int \frac{1}{\sqrt{x+1}} d x=2 \sqrt{x+1} \\
& (\text { put } x+1=t) \\
& 3 \int \frac{1}{(x-3) \sqrt{x+1}} d x=-\frac{3 \ln (\sqrt{x+1}+2)}{2}+\frac{3 \ln (\sqrt{x+1}-2)}{2} \\
& \left(\text { put } x+1=t^2\right) \\
& =-\frac{3 \ln (\sqrt{x+1}+2)}{2}+\frac{3 \ln (\sqrt{x+1}-2)}{2}+2 \sqrt{x+1} \\
& =2 \sqrt{x+1}+\frac{3}{2}\left(\log \frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right)+c
\end{aligned}$

Hence, the answer is the option 3.

Example 3: $\int \frac{d x}{(2 x+3) \sqrt{4 x+5}}=$
1) $\tan ^{-1} \sqrt{4 x-5}+c$
2) $4 \tan ^{-1} \sqrt{4 x+5}+c$
3) $\tan ^{-1} \sqrt{5 x+4}+c$
4) $\tan ^{-1} \sqrt{5 x-4}+c$

Solution
Let $I=\int \frac{d x}{(2 x+3) \sqrt{4 x+5}}$
Put $4 \mathrm{x}+5=\mathrm{t}^2$

$\therefore x=\frac{t^2-5}{4}$
and
$\begin{aligned}
& d x=2 t \cdot d t \\
& \therefore I=\int \frac{2 t \cdot d t}{\left(2\left(\frac{t^2-5}{4}\right)+3\right) \cdot t} \\
& \therefore I=\int \frac{4 d t}{\left(t^2+1\right)} \\
& \therefore I=4 \tan ^{-1}(t)+c \\
& \therefore I=4 \tan ^{-1} \sqrt{4 x+5}+c
\end{aligned}$

Hence, the answer is the option 2.

Example 4: The integral $\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}} \mathrm{~d} x$ is equal to : (where $C$ is a constant of integration)

1) $\frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{5}{4}}+C$
2) $\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C$
3) $\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{5}{4}}+C$
4) $\frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{1}{4}}+C$

Solution

$\begin{gathered}
I=\int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}} d x \\
I=\int \frac{1}{(x-1)^2\left(\frac{x+2}{x-1}\right)^{\frac{3}{4}}} d x
\end{gathered}$

Let $\frac{x+2}{x-1}=t \Rightarrow \frac{(x-1)-(x+2)}{(x-1)^2} d x=d t$

$\begin{aligned}
& \Rightarrow I=\frac{-1}{3} \int \frac{d t}{t^{\frac{5}{4}}}=\frac{-1}{3} \times \frac{t^{\frac{-5}{4}+1}}{-\frac{5}{4}+1}+c=\frac{4}{3} t^{\frac{-1}{4}}+c \\
& =\frac{4}{3} \times\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+c
\end{aligned}$
option (2)

Example 5: Let $I(x)=\int \sqrt{\frac{x+7}{x}} d x$ and $I(9)=12+7 \log _e 7$. If $I(1)=\alpha+7 \log _e(1+$ Example $5: 2 \sqrt{2}$ ), then $\alpha^4$ is equal to

1) $64$

2) $23$

3) $43$

4) $21$

Solution

$\begin{aligned}
& \int \sqrt{\frac{x+7}{x}} d x \\
& \text { Put } x=t^2 \\
& \mathrm{dx}=2 \mathrm{tdt} \\
& \int 2 \sqrt{\mathrm{t}^2+7} \mathrm{dt}=2 \int \sqrt{\mathrm{t}^2+\sqrt{7^2}} \mathrm{dt} \\
& \mathrm{I}(\mathrm{t})=2\left[\frac{\mathrm{t}}{2} \sqrt{\mathrm{t}^2+7}+\frac{7}{2} \ln \left|\sqrt{\mathrm{t}^2+7}\right|\right]+\mathrm{C} \\
& \mathrm{I}(\mathrm{x})=\sqrt{\mathrm{x}} \sqrt{\mathrm{x}+7}+7 \ln |\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}+7}|+\mathrm{C} \\
& \mathrm{I}(9)=12+7 \ln 7=12+7(\ln (3+4))+\mathrm{C} \\
& \Rightarrow \mathrm{C}=0 \\
& \mathrm{I}(\mathrm{x})=\sqrt{\mathrm{x}} \sqrt{\mathrm{x}+7}+7 \ln (\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}+7}) \\
& \mathrm{I}(1)=1 \sqrt{8}+7 \ln (1+\sqrt{8}) \\
& \mathrm{I}(1)=\sqrt{8}+7 \ln (1+2 \sqrt{2}) \\
& \alpha=\sqrt{8} \\
& \alpha^4=\left(8^{1 / 2}\right)^4 \\
& \alpha^4=8^2=64
\end{aligned}$

Hence, the answer is (64).

Frequently Asked Questions (FAQs)

Q: What is the role of complex numbers in integrating certain irrational functions?

Complex numbers can play a role in integrating certain irrational functions:

Q: How do you integrate expressions of the form (ax + b)^(1/n) where n is an integer greater than 2?

To integrate expressions of the form (ax + b)^(1/n) where n > 2:

Q: What is the significance of the substitution x = a sec θ in integration?

The substitution x = a sec θ is significant because:

Q: How do you approach integrals involving √(x/(1-x))?

To integrate expressions involving √(x/(1-x)):

Q: What is the strategy for integrating irrational functions with trigonometric terms?

For integrating irrational functions with trigonometric terms:

Q: How do you integrate expressions of the form √(ax² + bx + c)?

To integrate expressions of the form √(ax² + bx + c):

Q: What is the importance of recognizing perfect square trinomials under a square root?

Recognizing perfect square trinomials under a square root is important because:

Q: How do you handle integrals involving both square roots and cube roots?

For integrals involving both square roots and cube roots:

Q: What is the strategy for integrating (x² - a²)^(3/2)?

To integrate (x² - a²)^(3/2):

Q: How does the integration process change when dealing with complex irrational functions?

When dealing with complex irrational functions:

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