The value of ∫ (ex(x2 tan -1 x + tan-1 x+1)/x2 + 1) dx is equal to

Question

The value of ∫ (ex(x2 tan -1 x + tan-1 x+1)/x2 + 1) dx is equal to (A) ex tan-1x + c (B) - tan-1x(ex) + c (C)  tan-1(xe) + c (D) e tan-1x + c

Tardigrade Admin · Accepted Answer

Let I=∫ ex((x2 tan -1 x+ tan -1 x+1/x2+1)) d x ⇒ I=∫ ex( tan -1 x+(1/x2+1)) d x If f(x)= tan -1 x, then f'(x)=(1/x2+1) ∴ I=∫ ex( tan -1 x+(1/x2+1)) d x =ex tan -1 x+C [∵ ∫ ex[f(x)+f'(x)] d x=ex f(x)+C]

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

$e^{x} tan^{- 1} x + c$

$- tan^{- 1} x (e^{x}) + c$

$tan^{- 1} (x^{e}) + c$

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

Q. The value of ∫x2+1ex(x2tan−1x+tan−1x+1)​dx is equal to

extan−1x+c

−tan−1x(ex)+c

tan−1(xe)+c

etan−1x+c

Let I=∫ex(x2+1x2tan−1x+tan−1x+1​)dx ⇒I=∫ex(tan−1x+x2+11​)dx If f(x)=tan−1x, then f′(x)=x2+11​ ∴I=∫ex(tan−1x+x2+11​)dx =extan−1x+C [∵∫ex[f(x)+f′(x)]dx=exf(x)+C]

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

$e^{x} tan^{- 1} x + c$

$- tan^{- 1} x (e^{x}) + c$

$tan^{- 1} (x^{e}) + c$

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