#### Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 34 maths.

Hint: Use indefinite integral formula and the given limits to solve this integral.

Given: $\int_{0}^{1}\frac{24x^2}{(1+x^2)^4}dx$

Solution: $\int_{0}^{1}\frac{24x^2}{(1+x^2)^4}dx$

put  $(1+x^2)=t$  than, $2x \; dx = dt$

when x=0 then t=1 & when x=1 then t=2 ,

\begin{aligned} &\int_{0}^{1} \frac{24 x^{3}}{\left(1+x^{2}\right)^{4}} d x=12 \int_{1}^{2} \frac{(t-1)}{(t)^{4}} d t \\ &=12 \int_{1}^{2}\left(\frac{t}{(t)^{4}}-\frac{1}{(t)^{4}}\right) d t \\ &=12 \int_{1}^{2}\left(\frac{1}{(t)^{3}}-\frac{1}{(t)^{4}}\right) d t \\ &=12 \int_{1}^{2}(t)^{-3} d t-12 \int_{1}^{2}(t)^{-4} d t \end{aligned}

\begin{aligned} &=12\left[\frac{t^{-3+1}}{-3+1}\right]_{1}^{2}-12\left[\frac{t^{-4+1}}{-4+1}\right]_{1}^{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=12\left[\frac{t^{-2}}{-2}\right]_{1}^{2}-12\left[\frac{t^{-3}}{-3}\right]_{1}^{2} \\ &=\frac{-12}{2}\left[\frac{1}{t^{2}}\right]_{1}^{2}-\frac{12}{(-3)}\left[\frac{1}{t^{3}}\right]_{1}^{2} \end{aligned}

\begin{aligned} &=-6\left[\frac{1}{2^{2}}-1\right]+4\left[\frac{1}{2^{3}}-\frac{1}{1^{3}}\right] \\ &=-6\left[\frac{1}{4}-1\right]+4\left[\frac{1}{8}-1\right] \\ &=-6\left[\frac{1-4}{4}\right]+4\left[\frac{1-8}{8}\right] \\ &=-6\left[\frac{-3}{4}\right]+4\left[\frac{-7}{8}\right] \\ &=\frac{9}{2}-\frac{7}{2} \\ &=\frac{9-7}{2} \\ &=\frac{2}{2}=1 \end{aligned}