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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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}} = $

Similar Questions

If $\int \frac{(x-1) dx}{(x+1) \sqrt{x^3+x^2+x}} = A \cdot \tan^{-1} \sqrt{f(x)} + \text{constant}$, then the ordered pair $(A, f(-1)) =$

If $\int \frac{5 \tan x}{\tan x-2} d x = \alpha x + \beta \log |\sin x - 2 \cos x| + \gamma$,then $\alpha - \beta =$

If $\int(x+2) \sqrt{x^2-x+2} \, dx = \frac{1}{3} f(x) + \frac{5}{8} g(x) + \frac{35}{16} h(x) + c$, then $f(-1) + g(-1) + h\left(\frac{1}{2}\right) = $

For $0 < x < 1$,evaluate the integral $\int [\operatorname{Tan}^{-1}(1-x+x^2) + \operatorname{Tan}^{-1}(1-x)] dx$.

If $\frac{3 \pi}{2} < x < \frac{5 \pi}{2}$ and $\int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) \, dx = f(x) + c$ where $c$ is the constant of integration, then $f\left(\frac{\pi}{3}\right) - f(0) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module