Question Number 9107 by jainamanj98@gmail.com last updated on 18/Nov/16
Answered by FilupSmith last updated on 19/Nov/16
![(1) ∫_1 ^3 (x+3(√x))dx=∫_1 ^3 xdx+3∫_1 ^3 x^(1/2) dx =(1/2)[x^2 ]_1 ^3 +2[x^(3/2) ]_1 ^3 =4+2((√(27))−1) =2+2(√(27)) =2+2(√((3×9))) =2+6(√3) (2) ∫(x^3 /(^3 (√(x^4 +1))))dx u=x^4 +1 du=4x^3 ⇒(1/4)∫(1/u^(1/3) )du =(1/4)∫u^(−1/3) du =(1/4)((3/2)u^(2/3) )+c =(3/8)(x^4 +1)^(2/3) +c](https://www.tinkutara.com/question/Q9112.png)
$$\left(\mathrm{1}\right) \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} \left({x}+\mathrm{3}\sqrt{{x}}\right){dx}=\int_{\mathrm{1}} ^{\mathrm{3}} {xdx}+\mathrm{3}\int_{\mathrm{1}} ^{\mathrm{3}} {x}^{\mathrm{1}/\mathrm{2}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{x}^{\mathrm{2}} \right]_{\mathrm{1}} ^{\mathrm{3}} +\mathrm{2}\left[{x}^{\mathrm{3}/\mathrm{2}} \right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$=\mathrm{4}+\mathrm{2}\left(\sqrt{\mathrm{27}}−\mathrm{1}\right) \\ $$$$=\mathrm{2}+\mathrm{2}\sqrt{\mathrm{27}} \\ $$$$=\mathrm{2}+\mathrm{2}\sqrt{\left(\mathrm{3}×\mathrm{9}\right)} \\ $$$$=\mathrm{2}+\mathrm{6}\sqrt{\mathrm{3}} \\ $$$$\: \\ $$$$\left(\mathrm{2}\right) \\ $$$$\int\frac{{x}^{\mathrm{3}} }{\:^{\mathrm{3}} \sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}{dx} \\ $$$${u}={x}^{\mathrm{4}} +\mathrm{1} \\ $$$${du}=\mathrm{4}{x}^{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{1}}{{u}^{\mathrm{1}/\mathrm{3}} }{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int{u}^{−\mathrm{1}/\mathrm{3}} {du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}}{\mathrm{2}}{u}^{\mathrm{2}/\mathrm{3}} \right)+{c} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}/\mathrm{3}} +{c} \\ $$