Question

integral of tan inverse root x dx

Answer 1

Answer:

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Step-by-step explanation:

By Parts , u = tan⁻¹  (√x) , v= 1

∫ tan⁻¹  (√x) dx = tan⁻¹  (√x)   *  x   -  ∫  (1/(x+1) 1/2(√x)  x

= x*  tan⁻¹  (√x) - 1/2 ∫  (√x) /(x+1) dx  ,  

∫  (√x) /(x+1) dx   =  ∫  t / (t^2 +1)  2t dt = 2  ∫ [1 - 1/(1+t^2) dt] = 2t - 2  tan⁻¹ (√x)  

∫ tan⁻¹  (√x) dx =   x*  tan⁻¹  (√x) - 1/2 ∫  (√x) /(x+1) dx  = x*  tan⁻¹  (√x)  - (√x)  +  tan⁻¹ (√x)  

(x+1) tan⁻¹  (√x)  -  (√x)