int frac{1}{7-6 x-x^2} d x= | vedclass

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Question

(B) Let $I = \int \frac{1}{7-6 x-x^2} dx$. First,complete the square in the denominator: $7-6x-x^2 = 7 - (x^2+6x) = 7 - (x^2+6x+9-9) = 16 - (x+3)^2$.

Vedclass · Accepted Answer

$\frac{1}{8} \log \left(\frac{7+x}{1-x}\right)+c$,where $c$ is a constant of integration.

Vedclass · Answer

(B) Let $I = \int \frac{1}{7-6 x-x^2} dx$. First,complete the square in the denominator: $7-6x-x^2 = 7 - (x^2+6x) = 7 - (x^2+6x+9-9) = 16 - (x+3)^2$. So,$I = \int \frac{1}{16-(x+3)^2} dx$. Using the standard integral formula $\int \frac{1}{a^2-x^2} dx = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + c$,where $a=4$ and $x$ is replaced by $(x+3)$: $I = \frac{1}{2(4)} \log \left| \frac{4+(x+3)}{4-(x+3)} \right| + c$. $I = \frac{1}{8} \log \left| \frac{7+x}{1-x} \right| + c$.

$\int \frac{1}{7-6 x-x^2} d x=$

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