₁^{4} {x²+x}{{2x+1}} \, dx is equal to?

Correct Answer is : 57−35{57-{3}}{5}557−3

Correct Answer is : 57−35{57-{3}}{5}557−3

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UPSC NDA Math Model Paper 6

Question : 16 of 120

Marks: +1, -0

\int\limits_{1}^{4} \frac{x^{2}+x}{\sqrt{2x+1}} \, dx

is equal to?

Solution:

I=\int\limits_{1}^{4} \frac{x^{2}+x}{\sqrt{2x+1}} \, dx

Let

2x + 1 = t^2

..... (1)

Differentiaiting with respect to x, we get

⇒ 2dx = 2tdt

⇒ dx = tdt

From equation (1), we get

x=\frac{t^{2}-1}{2}

Now,

I=\int\limits_{\sqrt{3}}^{3} \frac{ \left( \frac{t^{2}-1}{2} \right)^{2} + \frac{t^{2}-1}{2} }{ \sqrt{t^{2}} } t \, dt

=\int\limits_{\sqrt{3}}^{3} \left( \frac{t^{4}-2t^{2}+1}{4} + \frac{t^{2}-1}{2} \right) \, dt

=\int\limits_{\sqrt{3}}^{3} \left( \frac{t^{4}-2t^{2}+1+2t^{2}-2}{4} \right) \, dt

=\frac{1}{4} \int\limits_{\sqrt{3}}^{3} (t^{4}-1) \, dt

=\frac{1}{4} \left[ \frac{t^{5}}{5} - t \right]_{\sqrt{3}}^{3}

=\frac{57-\sqrt{3}}{5}

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