int_0^1 frac{dx}{(3x+2)+sqrt{3x+2}} = . . . | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I = \int_0^1 \frac{dx}{(3x+2)+\sqrt{3x+2}}$.Substitute $u = \sqrt{3x+2}$. Then $u^2 = 3x+2$,so $2u du = 3 dx$,which means $dx = \frac{2}{3} u

Vedclass · Accepted Answer

$\frac{2}{3} \log \left|\frac{\sqrt{5}+1}{\sqrt{2}+1}\right|$

Vedclass · Answer

(C) Let $I = \int_0^1 \frac{dx}{(3x+2)+\sqrt{3x+2}}$.Substitute $u = \sqrt{3x+2}$. Then $u^2 = 3x+2$,so $2u du = 3 dx$,which means $dx = \frac{2}{3} u du$.When $x=0$,$u = \sqrt{2}$. When $x=1$,$u = \sqrt{5}$.Substituting these into the integral:$I = \int_{\sqrt{2}}^{\sqrt{5}} \frac{\frac{2}{3} u du}{u^2+u} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{u du}{u(u+1)} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{du}{u+1}$.Integrating,we get $I = \frac{2}{3} [\log |u+1|]_{\sqrt{2}}^{\sqrt{5}}$.$I = \frac{2}{3} (\log |\sqrt{5}+1| - \log |\sqrt{2}+1|) = \frac{2}{3} \log \left|\frac{\sqrt{5}+1}{\sqrt{2}+1}\right|$.Thus,the correct option is $C$.

$\int_0^1 \frac{dx}{(3x+2)+\sqrt{3x+2}} = $ . . . . . . .

Similar Questions

If $k \int_{0}^{1} x \cdot f(3x) \, dx = \int_{0}^{3} t \cdot f(t) \, dt$,then the value of $k$ is

The value of the definite integral,$\int\limits_0^{\sqrt {\ln \left( {\frac{\pi }{2}} \right)} } {\cos \left( {{e^{{x^2}}}} \right)} \cdot 2x {e^{{x^2}}}dx$ is

If $\alpha = \int_0^1 \left(e^{9x + 3 \tan^{-1} x}\right) \left(\frac{12 + 9x^2}{1 + x^2}\right) dx$,where $\tan^{-1} x$ takes only principal values,then the value of $\left(\log_e |1 + \alpha| - \frac{3\pi}{4}\right)$ is

The value of the integral $\int_0^{\log 5} {\frac{{{e^x}\sqrt {{e^x} - 1} }}{{{e^x} + 3}}} \,dx = $

If $\int_{0}^{1} f(x) dx = 5$,then the value of $\int_{0}^{1} f(x) dx + 100 \int_{0}^{1} x^{9} f(x^{10}) dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module