#### Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 28

$\frac{1}{2756}$

Hint:

Integrate then use limits

Given:

$\int_{0}^{1} \frac{x}{(1-x)^{54}} d x$

Solution:

\begin{aligned} &I=\int_{0}^{1} \frac{x}{(1-x)^{54}} d x \\\\ &I=\int_{0}^{1} \frac{x}{(x-1)^{54}} \quad\left[\mathrm{Q}(x-y)^{2}=(y-x)^{2}\right] \end{aligned}

\begin{aligned} &=\int_{0}^{1} \frac{x+1-1}{(x-1)^{54}} d x \\\\ &=\int_{0}^{1} \frac{(x-1)}{(x-1)^{54}} d x+\int_{0}^{1} \frac{1}{(x-1)^{54}} d x \end{aligned}

\begin{aligned} &=\int_{0}^{1}(x-1)^{-53} d x+\int_{0}^{1}(x-1)^{-54} d x \\\\ &=\left[-\frac{(x-1)^{-52}}{52}\right]_{0}^{1}+\left[-\frac{(x-1)^{-53}}{53}\right]_{0}^{1} \end{aligned}

\begin{aligned} &=\left[\frac{(x-1)^{-52}}{52}\right]_{0}^{1}+\left[\frac{(x-1)^{-53}}{53}\right]_{0}^{1} \\\\ &=\frac{(-1)^{52}}{52}+\frac{(-1)^{53}}{53} \end{aligned}

\begin{aligned} &=\frac{1}{52}-\frac{1}{53}=\frac{1}{52 \times 53} \\\\ &=\frac{1}{2756} \end{aligned}