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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18 .14 Question 10

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.14 Question 10

Answers (1)

Answer:\frac{x^{3}}{3}-x+2 \tan ^{-1}(x)+c

Hint: To solve this integral, use special integral formula.

Given: \int \frac{x^{4}+1}{x^{2}+1} d x

Solution:

Let

\begin{aligned} &I=\int \frac{x^{4}+1}{x^{2}+1} d x=\int \frac{x^{4}+1+2 x^{2}-2 x^{2}}{x^{2}+1} d x \\\\ &=\int \frac{\left\{\left(x^{2}\right)^{2}+1+2 x^{2}\right\}-2 x^{2}}{x^{2}+1} d x \quad\left[\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right] \end{aligned}

=\int \frac{\left(x^{2}+1\right)^{2}-2 x^{2}}{x^{2}+1} d x=\int \frac{\left(x^{2}+1\right)^{2}}{x^{2}+1}-\frac{2 x^{2}}{x^{2}+1} d x

 

 

\begin{aligned} &=\int\left\{\left(x^{2}+1\right)-\frac{\left[2 x^{2}+2-2\right]}{x^{2}+1}\right\} d x \\\\ &=\int\left\{\left(x^{2}+1\right)-\frac{\left[2\left(x^{2}+1\right)-2\right]}{x^{2}+1}\right\} d x \\\\ &=\int\left\{\left(x^{2}+1\right)-\frac{2\left(x^{2}+1\right)}{x^{2}+1}+\frac{2}{x^{2}+1}\right\} d x \end{aligned}

\begin{aligned} &=\int\left\{\left(x^{2}+1\right)-2+\frac{2}{x^{2}+1}\right\} d x \\\\ &=\int\left(x^{2}+1\right) d x-2 \int 1 d x+2 \int \frac{1}{x^{2}+1} d x \\ \\&=\int x^{2} d x+\int 1 d x-2 \int 1 d x+2 \int \frac{1}{x^{2}+1} d x \end{aligned}

 

 

=\frac{x^{2+1}}{2+1}+x-2 x+2 \cdot \frac{1}{1} \tan ^{-1}\left(\frac{x}{1}\right)+c \quad \quad \quad \quad \quad \quad\left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c, \int 1 d x=x+c \\\\ \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \end{array}\right]
=\frac{x^{3}}{3}-x+2 \tan ^{-1}(x)+c

 

Posted by

infoexpert27

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