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

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

Question

(B) To solve the integral $I = \int \frac{1}{(x+2) \sqrt{x^2+x+2}} \, dx$,we use the substitution $x+2 = \frac{1}{t}$.Then $dx = -\frac{1}{t^2} \, dt$

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1} \left( \frac{x+6}{\sqrt{7}(x+2)} \right) + c$

Vedclass · Answer

(B) To solve the integral $I = \int \frac{1}{(x+2) \sqrt{x^2+x+2}} \, dx$,we use the substitution $x+2 = \frac{1}{t}$.Then $dx = -\frac{1}{t^2} \, dt$.Substituting these into the integral:$I = \int \frac{1}{(1/t) \sqrt{(1/t - 2)^2 + (1/t - 2) + 2}} \cdot (-1/t^2) \, dt$$I = -\int \frac{1}{t \sqrt{1/t^2 - 4/t + 4 + 1/t - 2 + 2}} \, dt$$I = -\int \frac{1}{\sqrt{1 - 3t + 4t^2}} \, dt$$I = -\frac{1}{2} \int \frac{1}{\sqrt{t^2 - \frac{3}{4}t + \frac{1}{4}}} \, dt$Completing the square: $t^2 - \frac{3}{4}t + \frac{1}{4} = (t - \frac{3}{8})^2 + \frac{1}{4} - \frac{9}{64} = (t - \frac{3}{8})^2 + \frac{7}{64}$.Using the formula $\int \frac{dx}{\sqrt{x^2+a^2}} = \ln|x + \sqrt{x^2+a^2}|$,we get:$I = -\frac{1}{2} \ln \left| (t - \frac{3}{8}) + \sqrt{(t - \frac{3}{8})^2 + \frac{7}{64}} \right| + c$.Substituting $t = \frac{1}{x+2}$ back,we obtain the result in terms of $\operatorname{Sinh}^{-1}$.

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

Similar Questions

If $\int \frac{a \cos x+3 \sin x}{5 \cos x+2 \sin x} d x=\frac{26}{29} x-\frac{k}{29} \log |5 \cos x+2 \sin x|+c$,then $|a+k|=$

If $\int \frac{1}{x} \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=2 f(x)-2 \operatorname{Sin}^{-1} \sqrt{x}+c$,then $f(x)=$

Find the integral of the function $\frac{1}{\sin x \cos ^{3} x}$.

If $\int (e^{2x} + 2e^{x} - e^{-x} - 1) e^{(e^{x} + e^{-x})} dx = g(x) e^{(e^{x} + e^{-x})} + c$,where $c$ is a constant of integration,then $g(0)$ is equal to

$\int \sqrt{\frac{1-x}{1+x}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module