The value of int frac{d x}{7+6 x-x^2} is equa | vedclass

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(D) To evaluate the integral $I = \int \frac{d x}{7+6 x-x^2}$,we first complete the square for the quadratic expression in the denominator: $7+6 x-x^2

Vedclass · Accepted Answer

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

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(D) To evaluate the integral $I = \int \frac{d x}{7+6 x-x^2}$,we first complete the square for the quadratic expression in the denominator: $7+6 x-x^2 = -(x^2-6 x-7) = -(x^2-6 x+9-16) = 16-(x-3)^2$. Thus,the integral becomes $I = \int \frac{d x}{4^2-(x-3)^2}$. Using the standard formula $\int \frac{d x}{a^2-x^2} = \frac{1}{2 a} \log \left| \frac{a+x}{a-x} \right| + c$,where $a=4$ and $x$ is replaced by $(x-3)$: $I = \frac{1}{2(4)} \log \left| \frac{4+(x-3)}{4-(x-3)} \right| + c$. Simplifying the expression inside the logarithm: $I = \frac{1}{8} \log \left| \frac{x+1}{7-x} \right| + c$.

The value of $\int \frac{d x}{7+6 x-x^2}$ is equal to

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