∫ e tan -1 x(1+x+x2) ⋅ d( cot -1 x) is equal to

Question

∫ e tan -1 x(1+x+x2) ⋅ d( cot -1 x) is equal to (A) -e tan -1 x+C (B) e tan -1 x+C (C) -x ⋅ e tan -1 x+C (D) x ⋅ e tan -1 x+C

Tardigrade Admin · Accepted Answer

I =∫ e tan -1 x(1+x+x2) ⋅(-(1/1+x2)) d x =-∫ 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 =-∫ e tan -1 x d x-x ⋅ e tan -1 x+∫ e tan -1 x d x+ C =-x ⋅ e tan -1 x+C

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

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

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

$- x \cdot e^{t a n^{- 1} x} + C$

$x \cdot e^{t a n^{- 1} x} + C$

Q. ∫etan−1x(1+x+x2)⋅d(cot−1x) is equal to

−etan−1x+C

etan−1x+C

−x⋅etan−1x+C

x⋅etan−1x+C

I=∫etan−1x(1+x+x2)⋅(−1+x21​)dx =−∫etan−1x(1+1+x2x​)dx =−∫etan−1xdx−∫x⋅1+x2etan−1x​dx =−∫etan−1xdx−x⋅etan−1x+∫etan−1xdx+C =−x⋅etan−1x+C

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

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

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

$- x \cdot e^{t a n^{- 1} x} + C$

$x \cdot e^{t a n^{- 1} x} + C$