int frac{dx}{(1+x) sqrt{8+7x-x^2}} = | vedclass

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Question

(A) Let $I = \int \frac{dx}{(1+x) \sqrt{8+7x-x^2}}$.Factor the quadratic expression: $8+7x-x^2 = (8-x)(1+x)$.So, $I = \int \frac{dx}{(1+x) \sqrt{(8-x)

Vedclass · Accepted Answer

$-\frac{2}{9} \sqrt{\frac{8-x}{1+x}} + c$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{(1+x) \sqrt{8+7x-x^2}}$.Factor the quadratic expression: $8+7x-x^2 = (8-x)(1+x)$.So, $I = \int \frac{dx}{(1+x) \sqrt{(8-x)(1+x)}} = \int \frac{dx}{(1+x)^{3/2} \sqrt{8-x}} = \int \frac{dx}{(1+x)^2 \sqrt{\frac{8-x}{1+x}}}$.Let $t^2 = \frac{8-x}{1+x}$.Differentiating both sides with respect to $x$:$2t \frac{dt}{dx} = \frac{(1+x)(-1) - (8-x)(1)}{(1+x)^2} = \frac{-1-x-8+x}{(1+x)^2} = \frac{-9}{(1+x)^2}$.Thus, $\frac{dx}{(1+x)^2} = -\frac{2}{9} t \, dt$.Substituting these into the integral:$I = \int \frac{1}{t} \left(-\frac{2}{9} t \, dt\right) = -\frac{2}{9} \int dt = -\frac{2}{9} t + c$.Substituting $t = \sqrt{\frac{8-x}{1+x}}$ back:$I = -\frac{2}{9} \sqrt{\frac{8-x}{1+x}} + c$.

$\int \frac{dx}{(1+x) \sqrt{8+7x-x^2}} = $

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