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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)

best_answer

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

 

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