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#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.20 Question 9 Maths Textbook Solution.

Answer: $\frac{x^{5}}{5}-\frac{4 x^{3}}{3}+20 x-40 \tan ^{-1} \frac{x}{2}+c$

Hint: Using $\int \frac{1}{x} d x \text { and } \int \frac{1}{1+x^{2}} d x$

Explanation: Let

$I=\int \frac{x^{2}\left(x^{4}+4\right)}{x^{2}+4} d x$

$\frac{x^{2}\left(x^{4}+4\right)}{x^{2}+4}=\frac{\left(x^{2}+4-4\right)\left(x^{4}+4\right)}{x^{2}+4}=\frac{\left[\left(x^{2}+4\right)-4\right]\left(x^{4}+4\right)}{x^{2}+4}$

$=\frac{\left(x^{2}+4\right)\left(x^{4}+4\right)}{x^{2}+4}-\frac{4\left(x^{4}+4\right)}{x^{2}+4}$

$\therefore \int \frac{x^{2}\left(x^{4}+4\right)}{x^{2}+4} d x=\int\left(x^{4}+4\right) d x-4 \int \frac{x^{4}}{x^{2}+4} d x-16 \int \frac{1}{x^{2}+4} d x$

$=\int\left(x^{4}+4\right) d x-4 \int\left(x^{2}-\frac{4 x^{2}}{x^{2}+4}\right) d x-16 \int \frac{1}{x^{2}+4} d x$

$\left[\because \frac{x^{4}}{x^{2}+4}=x^{2}-\frac{4 x^{2}}{x^{2}+4}\right]$

$=\int\left(x^{4}+4\right) d x-4 \int x^{2} d x+16 \int \frac{x^{2}+4-4}{x^{2}+4} d x-16 \int \frac{1}{x^{2}+4} d x$

$=\int\left(x^{4}+4\right) d x-4 \int x^{2} d x+16 \int 1 d x-64 \int \frac{d x}{x^{2}+4}-16 \int \frac{d x}{x^{2}+4}$

$=\frac{x^{5}}{5}+4 x-\frac{4 x^{3}}{3}+16 x-\frac{64}{2} \tan ^{-1} \frac{x}{2}-\frac{16}{2} \tan ^{-1} \frac{x}{2}+c$

$=\frac{x^{5}}{5}+20 x-\frac{4 x^{3}}{3}-40 \tan ^{-1} \frac{x}{2}+c$

$=\frac{x^{5}}{5}-\frac{4 x^{3}}{3}+20 x-40 \tan ^{-1} \frac{x}{2}+c$

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