Integration 3

If ∫(x−1) dx(x+xx+x)x(x+1)=4tan⁡−1(g(x))+C \displaystyle \int \dfrac{ (x-1) \, dx}{ \left( x + x \sqrt x + \sqrt x\right) \sqrt {\sqrt x (x+1) }} = 4\tan^{-1}(g(x)) + C, where CC is an arbitrary constant of integration, find g2(1)g^2 (1) .

#Calculus

Note by Dhruv Joshi
3 years, 4 months ago

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Need help

Dhruv Joshi - 3 years, 4 months ago

First make the substitution : x=t  ⟹  dx=2tdt\sqrt{x}=t\implies dx=2tdt I=∫2(t2−1)dt(t2+t+1)t(t2+1)I= \int \dfrac{2(t^2-1)dt}{(t^2+t+1)\sqrt{t(t^2+1)}}

Divide numerator & denominator by t2t^2 :

I=∫2(1−1t2)dt(t+1t+1)t+1tI= \int \dfrac{2(1-\frac{1}{t^2})dt}{(t+\frac{1}{t}+1)\sqrt{t+\frac{1}{t}}}

Now , let t+1t=ut+\dfrac{1}{t}=u

I=∫2du(u+1)uI= \int \dfrac{2du}{(u+1)\sqrt{u}}

Now , the last one ! Let u=m  ⟹  du=(2m)dm\sqrt{u}=m \implies du =(2m)dm

I=∫(4)dmm2+1=4tan⁡−1m+CI= \int \dfrac{(4)dm}{m^2+1}=4\tan^{-1} m + C

Resubstituting , we get :

I=4tan⁡−1u+C=4tan⁡−1t+1t+C=4tan⁡−1x+1x+CI=4\tan^{-1} \sqrt{u} +C = 4\tan^{-1} \sqrt{t+\dfrac{1}{t}} +C = 4\tan^{-1} \sqrt{\sqrt{x}+\dfrac{1}{\sqrt{x}}} +C

A Former Brilliant Member - 3 years, 2 months ago
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