#### Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 3 Maths Textbook Solution.

Answer: $\frac{16}{3}$

Given: $\int_{1}^{5} \frac{x}{\sqrt{2 x-1}} d x$

Hint: Let the denominator$(2x-1) = t$

Solution:

$\int_{1}^{5} \frac{x}{\sqrt{2 x-1}} d x$

Let  $t=2 x-1=>x=\frac{t+1}{2}$

$dt = 2dx$                   (differentiating w.r.t  x)

\begin{aligned} &I=\int_{1}^{5} \frac{t+1}{\frac{2}{\sqrt{t}}} \times \frac{d t}{2} \\ & \end{aligned}

$I=\int_{1}^{5} \frac{t+1}{2 \sqrt{t}} d t=\frac{1}{4} \int_{1}^{5} \sqrt{t}+(t)^{\frac{-1}{2}} d t$

$=\frac{1}{4}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)_{1}^{5}$

$=\frac{1}{4}\left(\frac{2}{3}(2 x-1)^{\frac{3}{2}}+2(2 x-1)^{\frac{1}{2}}\right)_{1}^{5}$

\begin{aligned} &=\frac{1}{4}\left(\frac{2}{3}(27-1)+2(3-1)\right) \\ & \end{aligned}

$=\frac{1}{4}\left(\frac{2}{3}(26)+2(2)\right)$

\begin{aligned} &=\frac{1}{4}\left(\frac{52+12}{3}\right) \\ & \end{aligned}

$=\frac{1}{4} \times \frac{64}{3} \\$

$=\frac{16}{3}$