int_0^1 (1+x) log (1+x) , dx = | vedclass

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

Question

(D) Let $I = \int_0^1 (1+x) \log (1+x) \, dx$. Using integration by parts,let $u = \log(1+x)$ and $dv = (1+x) \, dx$. Then $du = \frac{1}{1+x} \, dx$

Vedclass · Accepted Answer

$\frac{-3}{4} + 2 \log 2$

Vedclass · Answer

(D) Let $I = \int_0^1 (1+x) \log (1+x) \, dx$. Using integration by parts,let $u = \log(1+x)$ and $dv = (1+x) \, dx$. Then $du = \frac{1}{1+x} \, dx$ and $v = \frac{(1+x)^2}{2}$. $I = \left[ \frac{(1+x)^2}{2} \log(1+x) \right]_0^1 - \int_0^1 \frac{(1+x)^2}{2} \cdot \frac{1}{1+x} \, dx$ $I = \left[ \frac{4}{2} \log 2 - 0 \right] - \frac{1}{2} \int_0^1 (1+x) \, dx$ $I = 2 \log 2 - \frac{1}{2} \left[ x + \frac{x^2}{2} \right]_0^1$ $I = 2 \log 2 - \frac{1}{2} \left( 1 + \frac{1}{2} \right) = 2 \log 2 - \frac{1}{2} \left( \frac{3}{2} \right)$ $I = 2 \log 2 - \frac{3}{4} = \frac{-3}{4} + 2 \log 2$.

$\int_0^1 (1+x) \log (1+x) \, dx =$

Similar Questions

The value of $\int_{0}^{4} \{ \sqrt{x} \} dx$,where $\{ \}$ denotes the fractional part of $x$,is

Dividing the interval $[0, 6]$ into $6$ equal parts and by using the trapezoidal rule,the value of $\int_0^6 x^3 \, dx$ is approximately:

$\int_a^b \frac{\log x}{x} \, dx = $

The value of the definite integral $\int\limits_0^{\frac{1}{{\sqrt 2 }}} {\frac{{{x^2}dx}}{{\sqrt {1 - {x^2}} \,(1 + \sqrt {1 - {x^2}} )}}} $ is

$\int_0^\pi \frac{1}{4+3 \cos x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module