text{If } int x[log (1+x)]^3 dx = frac{(1+x)^ | vedclass

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

Question

(D) Let $I = \int x[\log (1+x)]^3 dx$.Put $\log (1+x) = t$, so $1+x = e^t$ and $dx = e^t dt$.$I = \int (e^t - 1) t^3 e^t dt = \int (e^{2t} t^3 - e^t t

Vedclass · Accepted Answer

$6 \log (1+x) - 9(\log (1+x))^2 + 7(\log (1+x))^3 + C$

Vedclass · Answer

(D) Let $I = \int x[\log (1+x)]^3 dx$.Put $\log (1+x) = t$, so $1+x = e^t$ and $dx = e^t dt$.$I = \int (e^t - 1) t^3 e^t dt = \int (e^{2t} t^3 - e^t t^3) dt$.Using integration by parts, $\int e^{at} t^n dt = \frac{e^{at}}{a} [t^n - \frac{n t^{n-1}}{a} + \frac{n(n-1) t^{n-2}}{a^2} - \dots]$.$I = \frac{e^{2t}}{2} [t^3 - \frac{3t^2}{2} + \frac{6t}{4} - \frac{6}{8}] - e^t [t^3 - 3t^2 + 6t - 6] + C$.Substituting $e^t = 1+x$ and $t = \log (1+x)$:$I = \frac{(1+x)^2}{16} [8t^3 - 12t^2 + 12t - 6] - (1+x) [t^3 - 3t^2 + 6t - 6] + C$.Comparing with $\frac{(1+x)^2}{16} f(x) + (1+x) g(x)$, we get $f(x) = 8t^3 - 12t^2 + 12t - 6$ and $g(x) = -(t^3 - 3t^2 + 6t - 6) = -t^3 + 3t^2 - 6t + 6$.Thus, $f(x) + g(x) = 7t^3 - 9t^2 + 6t + C = 7(\log (1+x))^3 - 9(\log (1+x))^2 + 6 \log (1+x) + C$.

$\text{If } \int x[\log (1+x)]^3 dx = \frac{(1+x)^2}{16}(f(x)) + (1+x)(g(x)), \text{ then } f(x) + g(x) = $

Similar Questions

$\int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x}$ is equal to

$\int \sqrt{\frac{1+x}{1-x}} \, dx = $

Integrate the function: $\frac{1}{\cos (x+a) \cos (x+b)}$

Integrate the function: $\sqrt{x^{2}+4x-5}$

$\int e^{\sin x} \frac{(x \cos^3 x - \sin x)}{\cos^2 x} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module