int frac{d x}{(x+1) sqrt{x^2+4}} = | vedclass

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

Question

(A) ધારો કે $I = \int \frac{d x}{(x+1) \sqrt{x^2+4}}$.$x+1 = \frac{1}{t}$ આદેશ લેતા,$x = \frac{1}{t} - 1 = \frac{1-t}{t}$ અને $dx = -\frac{1}{t^2} dt$

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{5}} \sinh ^{-1}\left(\frac{4-x}{2(x+1)}\right)+c$

Vedclass · Answer

(A) ધારો કે $I = \int \frac{d x}{(x+1) \sqrt{x^2+4}}$.$x+1 = \frac{1}{t}$ આદેશ લેતા,$x = \frac{1}{t} - 1 = \frac{1-t}{t}$ અને $dx = -\frac{1}{t^2} dt$ મળે.આ કિંમતો સંકલનમાં મૂકતા:$I = \int \frac{-\frac{1}{t^2} dt}{\frac{1}{t} \sqrt{(\frac{1-t}{t})^2 + 4}} = -\int \frac{dt}{t \sqrt{\frac{1-2t+t^2+4t^2}{t^2}}} = -\int \frac{dt}{\sqrt{5t^2-2t+1}}$.છેદમાં પૂર્ણવર્ગ બનાવતા:$5t^2 - 2t + 1 = 5(t^2 - \frac{2}{5}t + \frac{1}{5}) = 5((t-\frac{1}{5})^2 + \frac{4}{25}) = 5(t-\frac{1}{5})^2 + \frac{4}{5}$.$I = -\int \frac{dt}{\sqrt{5(t-\frac{1}{5})^2 + \frac{4}{5}}} = -\frac{1}{\sqrt{5}} \int \frac{dt}{\sqrt{(t-\frac{1}{5})^2 + (\frac{2}{5})^2}}$.સૂત્ર $\int \frac{du}{\sqrt{u^2+a^2}} = \sinh^{-1}(\frac{u}{a})$ નો ઉપયોગ કરતા:$I = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{t-\frac{1}{5}}{2/5}) + C = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{5t-1}{2}) + C$.$t = \frac{1}{x+1}$ પાછા મૂકતા:$I = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{\frac{5}{x+1}-1}{2}) + C = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{5-x-1}{2(x+1)}) + C = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{4-x}{2(x+1)}) + C$.

$\int \frac{d x}{(x+1) \sqrt{x^2+4}} = $

Similar Questions

$\int \frac{3x+2}{4x^2+4x+5} dx = A \log(4x^2+4x+5) + B \tan^{-1}\left(\frac{2x+1}{2}\right) + c$,હોય તો $A+B=$

$\int \frac{25 x^2+8}{\sqrt{25 x^2+9}} d x=$

$\int \frac{d x}{\sin x+\cos x}=$

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

$\int \frac{1}{1 + \sin^2 x} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module