Question Number 15022 by Tinkutara last updated on 07/Jun/17
$$\mathrm{Evaluate}:\:\int_{\mathrm{1}} ^{\mathrm{4}} \frac{{x}^{\mathrm{2}} \:+\:{x}}{\:\sqrt{\mathrm{2}{x}\:+\:\mathrm{1}}}\:{dx}\:\left(\mathrm{Question}\:\mathrm{ID}:\right. \\ $$$$\left.\mathrm{53}\right)\:\mathrm{How}\:\mathrm{does}\:\mathrm{the}\:\mathrm{limits}\:\mathrm{change}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{solution}\:\mathrm{of}\:\mathrm{Q}.\:\mathrm{No}.\:\mathrm{53}? \\ $$
Answered by Joel577 last updated on 07/Jun/17
![Let u = 2x + 1 ⇔ x = ((u − 1)/2) du = 2 dx ∫_1 ^4 ((x(x + 1))/( (√(2x + 1)))) dx = ∫_1 ^4 ((((u − 1)/2) . ((u + 1)/2))/(2(√u))) du = (1/8) ∫_1 ^4 ((u^2 − 1)/( (√u))) du = (1/8) ∫_1 ^4 (u^2 − 1)(u^(−1/2) ) du = (1/8) ∫_1 ^4 u^(3/2) − u^(−1/2) du = (1/8) [(2/5)(2x + 1)^(5/2) − 2(2x + 1)^(1/2) ]_1 ^4 continue...](https://www.tinkutara.com/question/Q15040.png)
$$\mathrm{Let}\:{u}\:=\:\mathrm{2}{x}\:+\:\mathrm{1}\:\:\Leftrightarrow\:\:{x}\:=\:\frac{{u}\:−\:\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{du}\:=\:\mathrm{2}\:{dx} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\:\frac{{x}\left({x}\:+\:\mathrm{1}\right)}{\:\sqrt{\mathrm{2}{x}\:+\:\mathrm{1}}}\:{dx} \\ $$$$=\:\:\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\:\frac{\frac{{u}\:−\:\mathrm{1}}{\mathrm{2}}\:.\:\frac{{u}\:+\:\mathrm{1}}{\mathrm{2}}}{\mathrm{2}\sqrt{{u}}}\:\:{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{8}}\:\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\:\frac{{u}^{\mathrm{2}} \:−\:\mathrm{1}}{\:\sqrt{{u}}}\:{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{8}}\:\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\:\left({u}^{\mathrm{2}} \:−\:\mathrm{1}\right)\left({u}^{−\mathrm{1}/\mathrm{2}} \right)\:{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{8}}\:\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\:{u}^{\mathrm{3}/\mathrm{2}} \:−\:{u}^{−\mathrm{1}/\mathrm{2}} \:{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{8}}\:\left[\frac{\mathrm{2}}{\mathrm{5}}\left(\mathrm{2}{x}\:+\:\mathrm{1}\right)^{\mathrm{5}/\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{2}{x}\:+\:\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} \right]_{\mathrm{1}} ^{\mathrm{4}} \\ $$$${continue}… \\ $$