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Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 23

Answers (1)

Answer: $\frac{57}{5}-\sqrt{5}$

Given: $\int_{2}^{4} \frac{x^{2}+x}{\sqrt{2 x+x}} d x$

Hint: Use substitution method

Solution: $\int_{2}^{4} \frac{x^{2}+x}{\sqrt{2 x+x}} d x$

Let

$\begin{aligned} &2 x+1=t^{2} \\ & \end{aligned}$

$2 d x=2 t d t \\$ (Differentiate w.r.t to x)

$d x=t d t$

Now, $\int_{2}^{4} \frac{x^{2}+x}{\sqrt{2 x+x}} d x$

$\begin{aligned} &=\int_{2}^{4} \frac{\left(\frac{t^{2}-1}{2}\right)^{2}+\left(\frac{t^{2}-1}{2}\right)}{t} t d t \\ & \end{aligned}$

$=\int_{2}^{4} \frac{\left(t^{2}-1\right)^{2}+2\left(t^{2}-1\right)}{4} d t$

$\begin{aligned} &=\frac{1}{4} \int_{2}^{4}\left(t^{4}+1-2 t^{2}+2 t^{2}-2\right) d t \\ & \end{aligned}$

$=\frac{1}{4} \int_{2}^{4} t^{4}-1 d t \\$

$=\frac{1}{4}\left(\frac{t^{5}}{5}-t\right)^{4}$

$\begin{aligned} &=\frac{1}{4}\left[\frac{(2 x+1)^{2} \sqrt{2 x+1}}{5}-\sqrt{2 x+1}\right]_{2}^{4} \\ & \end{aligned}$

$=\frac{1}{4}\left[\left(\frac{81 \times 3}{5}-3\right)-\left(\frac{25 \sqrt{5}}{5}-\sqrt{5}\right)\right]$

$\begin{aligned} &=\frac{1}{4}\left[\left(\frac{243-15}{5}\right)-\left(\frac{25 \sqrt{5}-5 \sqrt{5}}{5}\right)\right] \\ & \end{aligned}$

$=\frac{1}{4} \times \frac{228}{5}-\frac{1}{4} \times \frac{20 \sqrt{5}}{5} \\$

$=\frac{57}{5}-\sqrt{5}$

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