Integration of Irrational Algebraic Function

Integration of Irrational Algebraic Functions

When you first look at an integral involving a square root or a fractional exponent, it often feels messy — almost like trying to untangle a pair of tightly knotted headphones. These are what we call irrational algebraic functions, and they show up in many real mathematical problems, from geometry to physics. In this section, we break them down, learn the exact substitutions needed, and walk through step-by-step solved examples so the entire process feels smooth and predictable.

What Are Irrational Functions?

Irrational functions are those in which the variable appears inside a radical or inside a term with a fractional exponent.
For example:

• $ \sqrt{x+3} $

• $ \sqrt{ax^2 + bx + c} $

• $ (x+1)^{\frac{1}{3}} $

Any function containing such expressions is considered an irrational algebraic function. Integrating them requires specific substitutions to simplify the radical and convert the integral into a standard form.

Integrals Involving Linear Expressions Inside Square Roots

These are integrals where the irrational part appears as $ \sqrt{px + q} $.

1. $ \int \frac{1}{(ax + b)\sqrt{px + q}} , dx $

2. $ \int \frac{ax + b}{\sqrt{px + q}} , dx $

3. $ \int \frac{\sqrt{px + q}}{ax + b} , dx $

4. $ \int \frac{1}{(ax^2 + bx + c)\sqrt{px + q}} , dx $

Standard Substitution

To simplify such integrals, use the substitution:
$ px + q = t^2 $
which gives
$ dx = \frac{2t}{p} , dt $

This removes the square root and transforms the integral into a rational function of $t$.

Solved Example:

Evaluate $ \int \frac{dx}{(x+1)\sqrt{x+2}} $

Let $ I = \int \frac{dx}{(x+1)\sqrt{x+2}} $

Put $ x + 2 = t^2 $
So, $ dx = 2t , dt $
and
$ x + 1 = t^2 - 1 $

Substitute:

$
I = \int \frac{2t , dt}{(t^2 - 1)\sqrt{t^2}}
$

Since $ \sqrt{t^2} = t $, the expression becomes:

$
I = 2 \int \frac{dt}{t^2 - 1}
$

Now integrate:

$
\int \frac{dt}{t^2 - 1} = \frac{1}{2} \ln \left| \frac{t - 1}{t + 1} \right| + C
$

So,

$
I = \ln \left| \frac{t - 1}{t + 1} \right| + C
$

Back-substitute $ t = \sqrt{x+2} $:

$
I = \ln \left| \frac{\sqrt{x+2} - 1}{\sqrt{x+2} + 1} \right| + C
$

Integrals Where the Linear Term Is in the Denominator

These integrals look tricky because they combine a linear expression in the denominator with a quadratic expression under a square root. Direct methods seldom work cleanly here, so a clever substitution becomes your best friend.

Form of the Integral

$ \int \frac{1}{(px + q)\sqrt{ax^2 + bx + c}} , dx $

The mix of a linear factor and a quadratic radical typically makes the integrand resistant to basic substitution or algebraic manipulation.

Standard Substitution to Use

Let $ px + q = \frac{1}{t} $

Then, $ x = \frac{1}{p}\left(\frac{1}{t} - q\right) $
and
$ dx = -\frac{1}{p t^2} , dt $

Once you substitute these into the integral, the linear denominator collapses beautifully into $t$, and the quadratic expression inside the square root transforms into a rational form involving $t$.

Why This Substitution Helps

This technique works well because:

• It flips the linear denominator, converting a difficult term into a simple $t$.

• The quadratic expression $ax^2 + bx + c$ becomes a rational function of $t$, removing the complexity of the square root.

• The resulting integral often simplifies to a standard form involving logarithms or inverse trigonometric functions.

Integrals Involving Quadratic Expressions Under Square Roots

When the integrand includes a quadratic expression inside a square root along with another quadratic (or linear) term in the denominator, the expression can feel a bit heavy at first glance. A great way to ease the complexity is by flipping the variable through substitution.

Form of the Integral

$ \int \frac{1}{(a x^2 + b)\sqrt{p x^2 + q}} , dx $

This structure usually combines a rational term with a square-root expression, making direct integration pretty tough.

Substitution to Use

Let
$ x = \frac{1}{t} $

Then,
$ dx = -\frac{1}{t^2} dt $

This substitution turns the expression inside the square root into:

$ \sqrt{p x^2 + q} = \sqrt{\frac{p}{t^2} + q} = \frac{\sqrt{p + q t^2}}{t} $

Suddenly, the integrand becomes a rational function of $t$, which is much easier to manage.

Why This Substitution Works

Using $ x = \frac{1}{t} $ is particularly effective when:

• The denominator involves $ax^2 + b$, which becomes $a/t^2 + b$—a more manageable rational form.

• The radical involves $px^2 + q$, which simplifies beautifully after inversion.

• The powers of $t$ line up neatly, allowing easy cancellations and resulting in a standard integral.

Step-by-Step Strategy for Irrational Integrals

This section walks you through a clear, methodical approach to breaking down irrational integrals, helping you choose the right substitution and simplify complex radicals with confidence.

1. Identify the Radical

Check whether the radical is linear, quadratic, or more complex.

2. Choose the Correct Substitution

• For $ \sqrt{px + q} $, use $ px + q = t^2 $

• For $ \sqrt{ax^2 + bx + c} $, consider completing the square

• For reciprocal forms, use $ x = \frac{1}{t} $

3. Simplify the Integral

Convert everything to $t$ and reduce the integral to a rational form.

4. Integrate Using Standard Formulas

Logarithmic or inverse trigonometric forms often appear.

5. Substitute Back

Always express the final answer in terms of the original variable.

Solved Examples Based On Integration Of Irrational Functions

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

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

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

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

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

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

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).

List of Topics Related to the Integration of Irrational Algebraic Functions

This section introduces key topics related to the integration of irrational algebraic functions, focusing on methods that simplify complex expressions involving radicals and fractional powers.

Application of Integrals

Integral of Particular Functions

Indefinite Integrals

Integration by parts

Application of Inequality in Definite Integration

NCERT Resources

This section gathers all the essential NCERT study materials for Chapter 7 – Integrals, providing a comprehensive resource to support your learning and exam preparation. It includes concise notes, detailed solutions, and exemplar problems that cover fundamental integration methods such as substitution, partial fractions, and integration by parts. With these materials, you can study systematically, strengthen your understanding, and boost your confidence for board exams and competitive tests.

NCERT Class 12 Maths Notes for Chapter 7 - Integrals

NCERT Class 12 Maths Solutions for Chapter 7 - Integrals

NCERT Class 12 Maths Exemplar Solutions for Chapter 7 - Integrals

Practice Questions based on Integration of Irrational Algebraic Function

This section provides targeted practice questions on integrating irrational algebraic functions, helping you strengthen your problem-solving skills and mastery through focused exercises. These questions are designed to build confidence and improve your application of integration techniques effectively.

Integration Of Irrational Algebraic Function- Practice Question MCQ

We have provided below the practice questions related to different concepts of integration to improve your understanding: