If int {ln ({x^2} + x)dx = xln ({x^2} + x) + | vedclass

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(D) We are given $\int {\ln ({x^2} + x)dx} = x\ln ({x^2} + x) + A$. Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$. L

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(D) We are given $\int {\ln ({x^2} + x)dx} = x\ln ({x^2} + x) + A$. Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = \ln(x^2+x)$ and $dv = dx$. Then $du = \frac{2x+1}{x^2+x} dx$ and $v = x$. So,$\int {\ln ({x^2} + x)dx} = x \ln(x^2+x) - \int x \cdot \frac{2x+1}{x^2+x} dx$. $= x \ln(x^2+x) - \int \frac{2x^2+x}{x^2+x} dx$. $= x \ln(x^2+x) - \int \frac{2(x^2+x) - x}{x^2+x} dx$. $= x \ln(x^2+x) - \int (2 - \frac{x}{x(x+1)}) dx$. $= x \ln(x^2+x) - \int (2 - \frac{1}{x+1}) dx$. $= x \ln(x^2+x) - (2x - \ln|x+1|) + C$. $= x \ln(x^2+x) - 2x + \ln|x+1| + C$. Comparing this with $x \ln(x^2+x) + A$,we get $A = -2x + \ln|x+1| + C$.

If $\int {\ln ({x^2} + x)dx = x\ln ({x^2} + x) + A}$,then $A = $

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