If ∫1x3+x4dx=Ax2+Bx+logx|x+1|+C, then 2A+2B=

We have, I=∫1x3+x4dx=∫1x3(x+1)dx=∫1x2(x(x+1))dx⇒I=∫1x21x−1x+1dx=∫1x1x2−1x(x+1)dx⇒I=∫1x1x2−1x−1x+1dx=∫1x3−1x2+1x(x+1)dx⇒I=∫1x3−1x2+1x−1x+1dx=−121x2+1x+log⁡xx+1+C∴A=−12 and B=1⇒2A+2B=1

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If $\int \frac{1}{x^{3} + x^{4}} d x = \frac{A}{x^{2}} + \frac{B}{x} + \log \frac{x}{| x + 1 |} + C,$ then $2 A + 2 B =$

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Detailed Solution

We have,

$\begin{array}{l} I = \int \frac{1}{x^{3} + x^{4}} d x = \int \frac{1}{x^{3} (x + 1)} d x = \int \frac{1}{x^{2} (x (x + 1))} d x \\ \Rightarrow I = \int \frac{1}{x^{2}} (\frac{1}{x} - \frac{1}{x + 1}) d x = \int \frac{1}{x} \{\frac{1}{x^{2}} - \frac{1}{x (x + 1)}\} d x \\ \Rightarrow I = \int \frac{1}{x} \{\frac{1}{x^{2}} - (\frac{1}{x} - \frac{1}{x + 1})\} d x = \int \frac{1}{x^{3}} - \frac{1}{x^{2}} + \frac{1}{x (x + 1)} d x \\ \Rightarrow I = \int (\frac{1}{x^{3}} - \frac{1}{x^{2}} + \frac{1}{x} - \frac{1}{x + 1}) d x = - \frac{1}{2} \frac{1}{x^{2}} + \frac{1}{x} + \log |\frac{x}{x + 1}| + C \end{array}$