If int tan^{-1} x , dx = Ax tan^{-1} x + B lo | vedclass

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(B) Using integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = 	an^{-1} x$ and $dv = dx$. Then $du = \frac{1}{1+x^2} dx$ and $v = x$. T

Vedclass · Accepted Answer

$1/2$

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(B) Using integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = 	an^{-1} x$ and $dv = dx$. Then $du = \frac{1}{1+x^2} dx$ and $v = x$. The integral becomes $x 	an^{-1} x - \int \frac{x}{1+x^2} dx$. To solve $\int \frac{x}{1+x^2} dx$, let $t = 1+x^2$, so $dt = 2x \, dx$, which means $x \, dx = \frac{1}{2} dt$. Thus, $\int \frac{x}{1+x^2} dx = \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \log(1+x^2)$. Substituting this back, we get $x 	an^{-1} x - \frac{1}{2} \log(1+x^2) + C$. Comparing this with $Ax 	an^{-1} x + B \log(1+x^2) + C$, we get $A = 1$ and $B = -1/2$. Therefore, $A + B = 1 - 1/2 = 1/2$.

If $\int \tan^{-1} x \, dx = Ax \tan^{-1} x + B \log(1 + x^2) + C$, then $A + B = \_\_\_\_$

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