∫ e tan-1x (1 + (x/1+x2))dx is equal to

Question

∫ e tan-1x (1 + (x/1+x2))dx is equal to (A)  (1/2)x e tan-1x+c (B) x e tan-1x+c (C) e tan-1x+c (D)  (1/2)e tan-1x+c

Tardigrade Admin · Accepted Answer

Let I=∫ e tan -1 x(1+(x/1+x2)) d x =∫ e tan -1 x d x+∫ (x e tan -1 x/1+x2) d x =x e tan -1 x-∫ (x e tan -1 x/1+x2) d x+∫ (x e tan -1 x/1+x2) d x+c =x e tan -1 x+c

Q. $\int e^{t a n - 1 x} (1 + \frac{x}{1 + x ^{2}}) d x$ is equal to

$\frac{1}{2} x e^{t a n^{- 1} x} + c$

$x e^{t a n^{- 1} x} + c$

$e^{t a n^{- 1} x} + c$

$\frac{1}{2} e^{t a n^{- 1} x} + c$

Q. ∫etan−1x(1+1+x2x​)dx is equal to

21​xetan−1x+c

xetan−1x+c

etan−1x+c

21​etan−1x+c

Let I=∫etan−1x(1+1+x2x​)dx =∫etan−1xdx+∫1+x2xetan−1x​dx =xetan−1x−∫1+x2xetan−1x​dx+∫1+x2xetan−1x​dx+c =xetan−1x+c

Q. $\int e^{t a n - 1 x} (1 + \frac{x}{1 + x ^{2}}) d x$ is equal to

$\frac{1}{2} x e^{t a n^{- 1} x} + c$

$x e^{t a n^{- 1} x} + c$

$e^{t a n^{- 1} x} + c$

$\frac{1}{2} e^{t a n^{- 1} x} + c$