Question

\(\rm \int_1^4 \frac{x^2 + x}{\sqrt{2x+1}}\;dx\) is equal to?
1. \(\frac{57-\sqrt 3}{5} \)
2. \(\frac{57-\sqrt 3}{4} \)
3. \(\frac{57-4\sqrt 3}{5} \)
4. None of the above

Answer 1

Correct Answer - Option 1 : \(\frac{57-\sqrt 3}{5} \)

Calculation:

I = \(\rm \int_1^4 \frac{x^2 + x}{\sqrt {2x+1}}\;dx\)

Let 2x + 1 = t²         ..... (1)

Differentiaiting with respect to x, we get

⇒ 2dx = 2tdt

⇒ dx = tdt

x	1	4
t	\(\sqrt 3\)	3

From equation (1), we get

x = \(\rm \frac{t^2-1}{2}\)

Now,

I = \(\rm \int_{\sqrt 3}^{3} \frac{(\rm \frac{t^2-1}{2})^2 + \rm \frac{t^2-1}{2}}{\sqrt{t^2}}\;tdt \)

= \(\rm \int_{\sqrt 3}^{3} \left(\frac{t^4-2t^2+1}{4}+ \rm \frac{t^2-1}{2} \right )dt\)

= \(\rm \int_{\sqrt 3}^{3} \left(\frac{t^4-2t^2+1+2t^2-2}{4} \right )dt\)

= \(\rm \frac{1}{4}\int_{\sqrt 3}^{3} (t^4-1)dt\)

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

= \(\frac{57-\sqrt 3}{5} \)