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

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Answer: \frac{16}{3}

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

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


\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{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}


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

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


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