If int frac{dx}{32-2x^2} = A log(4-x) + B log | vedclass

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(C) Given integral $I = \int \frac{dx}{32-2x^2} = \frac{1}{2} \int \frac{dx}{16-x^2}$.Using the formula $\int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \

Vedclass · Accepted Answer

$\frac{-1}{16}, \frac{1}{16}$

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(C) Given integral $I = \int \frac{dx}{32-2x^2} = \frac{1}{2} \int \frac{dx}{16-x^2}$.Using the formula $\int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + c$,where $a=4$:$I = \frac{1}{2} \left[ \frac{1}{2(4)} \log \left| \frac{4+x}{4-x} \right| \right] + c$$I = \frac{1}{16} [ \log |4+x| - \log |4-x| ] + c$$I = -\frac{1}{16} \log |4-x| + \frac{1}{16} \log |4+x| + c$.Comparing this with $A \log(4-x) + B \log(4+x) + c$,we get $A = -\frac{1}{16}$ and $B = \frac{1}{16}$.

If $\int \frac{dx}{32-2x^2} = A \log(4-x) + B \log(4+x) + c$,then the values of $A$ and $B$ are respectively (where $c$ is a constant of integration).

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