For x 0,if int (log x)^5 dx is equal to x[A | vedclass

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

Question

(A) Let $I = \int (\log x)^5 dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then $I = \int t^5 e^t dt$. Using integration by

Vedclass · Accepted Answer

$-44$

Vedclass · Answer

(A) Let $I = \int (\log x)^5 dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then $I = \int t^5 e^t dt$. Using integration by parts repeatedly,$\int t^n e^t dt = e^t [t^n - n t^{n-1} + n(n-1) t^{n-2} - \dots + (-1)^n n!]$. For $n = 5$,we have $I = e^t [t^5 - 5t^4 + 20t^3 - 60t^2 + 120t - 120] + C$. Substituting $t = \log x$ back,we get $I = x [(\log x)^5 - 5(\log x)^4 + 20(\log x)^3 - 60(\log x)^2 + 120(\log x) - 120] + C$. Comparing this with the given expression,we identify the coefficients: $A = 1, B = -5, C = 20, D = -60, E = 120, F = -120$. Thus,$A + B + C + D + E + F = 1 - 5 + 20 - 60 + 120 - 120 = -44$.

For $x > 0$,if $\int (\log x)^5 dx$ is equal to $x[A(\log x)^5 + B(\log x)^4 + C(\log x)^3 + D(\log x)^2 + E(\log x) + F] + \text{constant}$,then $A + B + C + D + E + F$ is equal to

Similar Questions

If $\int {\frac{{\sqrt {1 - {x^2}} }}{{{x^4}}}} dx\, = \,A(x)\,{(\sqrt {1 - {x^2}} )^m}\, + \,C,$ for a suitably chosen integer $m$ and a function $A(x),$ where $C$ is a constant of integration,then $(A(x))^m$ equals

If $\int e^u \sin 2x \, dx$ can be expressed in terms of elementary functions of $x$,then $u$ can be:

Integrate the function: $\frac{6x+7}{\sqrt{(x-5)(x-4)}}$

If $\int \frac{(x^2-1)}{(x+1)^2 \sqrt{x(x^2+x+1)}} dx = A \tan^{-1}\left(\sqrt{\frac{x^2+x+1}{x}}\right) + C$, where $C$ is a constant, then $A$ equals to

If $\int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} = (\tan x)^A + C(\tan x)^B + k$ where $k$ is a constant of integration,then $A+B+C$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module