Evaluate: _2^4 (x^2+x ) 2 x+1 d x - Vidyadip

MCQ

Evaluate: $\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x$

A
$57-5 \sqrt{5}$
B
$\frac{57-\sqrt{5}}{5}$
C
$\frac{57+5 \sqrt{5}}{5}$
✓
$\frac{57-5 \sqrt{5}}{5}$

✓

Answer

Correct option: D.

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

We have, $\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x$
Integrating by parts, we get
$\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x=\left[\left(x^2+x\right) \cdot \sqrt{2 x+1}\right]_2^4-\int_2^4(2 x+1) \cdot \sqrt{2 x+1} d x$
$=(60-6 \sqrt{5})-\int_2^4(2 x+1)^{3 / 2} d x$
$=(60-6 \sqrt{5})-\frac{1}{5} \cdot\left[(2 x+1)^{5 / 2}\right]_2^4$
$=(60-6 \sqrt{5})-\left(\frac{243}{5}-5 \sqrt{5}\right)$
$=\left(\frac{57}{5}-\sqrt{5}\right)=\left(\frac{57-5 \sqrt{5}}{5}\right)$

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