If I(x) = int x^2(log x)^2 dx and I(1) = 0,th | vedclass

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(C) We are given the integral $I(x) = \int x^2(\log x)^2 dx$. Using integration by parts,where $u = (\log x)^2$ and $dv = x^2 dx$,we have $du = \frac{

Vedclass · Accepted Answer

$\frac{x^3}{27}[9(\log x)^2 - 6 \log x + 2] - \frac{2}{27}$

Vedclass · Answer

(C) We are given the integral $I(x) = \int x^2(\log x)^2 dx$. Using integration by parts,where $u = (\log x)^2$ and $dv = x^2 dx$,we have $du = \frac{2 \log x}{x} dx$ and $v = \frac{x^3}{3}$.$I(x) = \frac{x^3}{3}(\log x)^2 - \int \frac{x^3}{3} \cdot \frac{2 \log x}{x} dx$$I(x) = \frac{x^3}{3}(\log x)^2 - \frac{2}{3} \int x^2 \log x dx$Applying integration by parts again for $\int x^2 \log x dx$ with $u = \log x$ and $dv = x^2 dx$:$\int x^2 \log x dx = \frac{x^3}{3} \log x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3} \log x - \frac{x^3}{9} + C_1$Substituting this back into the expression for $I(x)$:$I(x) = \frac{x^3}{3}(\log x)^2 - \frac{2}{3} [\frac{x^3}{3} \log x - \frac{x^3}{9}] + C$$I(x) = \frac{x^3}{3}(\log x)^2 - \frac{2x^3}{9} \log x + \frac{2x^3}{27} + C$$I(x) = \frac{x^3}{27} [9(\log x)^2 - 6 \log x + 2] + C$Given $I(1) = 0$,we substitute $x = 1$:$I(1) = \frac{1}{27} [9(0)^2 - 6(0) + 2] + C = 0$$\frac{2}{27} + C = 0 \Rightarrow C = -\frac{2}{27}$Thus,$I(x) = \frac{x^3}{27} [9(\log x)^2 - 6 \log x + 2] - \frac{2}{27}$.

If $I(x) = \int x^2(\log x)^2 dx$ and $I(1) = 0$,then $I(x)$ is equal to:

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