The value of I=int frac{x^2}{(a+b x)^2} ,d x | vedclass

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(C) Let $a+b x=t$. Then $x=\frac{t-a}{b}$ and $dx=\frac{dt}{b}$.Substituting these into the integral:$I = \int \frac{(\frac{t-a}{b})^2}{t^2} \cdot \fr

Vedclass · Accepted Answer

$\frac{1}{b^3}\left[a+b x-2 a \log |a+b x|-\frac{a^2}{a+b x}\right]+c$,(where $c$ is the constant of integration)

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(C) Let $a+b x=t$. Then $x=\frac{t-a}{b}$ and $dx=\frac{dt}{b}$.Substituting these into the integral:$I = \int \frac{(\frac{t-a}{b})^2}{t^2} \cdot \frac{dt}{b} = \frac{1}{b^3} \int \frac{t^2-2at+a^2}{t^2} dt$$I = \frac{1}{b^3} \int (1 - \frac{2a}{t} + \frac{a^2}{t^2}) dt$$I = \frac{1}{b^3} [t - 2a \log |t| - \frac{a^2}{t}] + c$Substituting $t=a+bx$ back:$I = \frac{1}{b^3} [a+bx - 2a \log |a+bx| - \frac{a^2}{a+bx}] + c$

The value of $I=\int \frac{x^2}{(a+b x)^2} \,d x$ is

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