Evaluate: ∫_1 ^4 ((x^2 + x)/(√(2x + 1))) dx (Question ID: 53) How does the limits change in the solution of Q. No. 53?


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Question Number 15022 by Tinkutara last updated on 07/Jun/17

$Evaluate : \int_{1}^{4} \frac{x^{2} + x}{\sqrt{2 x + 1}} d x (Question ID :$ $53) How does the limits change in the$ $solution of Q . No . 53 ?$
	Answered by Joel577 last updated on 07/Jun/17

	$Let u = 2 x + 1 \Leftrightarrow x = \frac{u - 1}{2}$ $d u = 2 d x$ $\underset{1}{\int^{4}} \frac{x (x + 1)}{\sqrt{2 x + 1}} d x$ $= \underset{1}{\int^{4}} \frac{\frac{u - 1}{2} . \frac{u + 1}{2}}{2 \sqrt{u}} d u$ $= \frac{1}{8} \underset{1}{\int^{4}} \frac{u^{2} - 1}{\sqrt{u}} d u$ $= \frac{1}{8} \underset{1}{\int^{4}} (u^{2} - 1) (u^{- 1 / 2}) d u$ $= \frac{1}{8} \underset{1}{\int^{4}} u^{3 / 2} - u^{- 1 / 2} d u$ $= \frac{1}{8} {[\frac{2}{5} {(2 x + 1)}^{5 / 2} - 2 {(2 x + 1)}^{1 / 2}]}_{1}^{4}$ $c o n t i n u e . . .$
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