The integral displaystyle ∫ tan- 1 x dx is equal to (where, C is the constant of integration)

Question

The integral displaystyle ∫ tan- 1 x dx is equal to (where, C is the constant of integration) (A) (1/1 + x2)+C (B) xtan- 1x+(1/2)log|. 1 + x2 |.+C (C) xtan- 1x+(1/2).(tan- 1 x/1 + x2)+C (D) xtan- 1x-(1/2)log|.1+x2|.+C

Tardigrade Admin · Accepted Answer

Let, I= displaystyle ∫ 1.tan- 1xdx =( tan -1 x) ⋅ x-∫ (1/1+x2) ⋅ x d x I=xtan- 1x-(1/2) displaystyle ∫ (2 x/1 + x2) d x I=xtan- 1x-(1/2)log|. 1 + x2 |.+C

Q. The integral $\int t a n^{- 1} x d x$ is equal to (where, $C$ is the constant of integration)

$\frac{1}{1 + x ^{2}} + C$

$x t a n^{- 1} x + \frac{1}{2} l o g ∣ 1 + x^{2} ∣ + C$

$x t a n^{- 1} x + \frac{1}{2} . \frac{t a n ^{- 1} x}{1 + x ^{2}} + C$

$x t a n^{- 1} x - \frac{1}{2} l o g ∣ 1 + x^{2} ∣ + C$

Let, $I = \int 1. t a n^{- 1} x d x$
$= (tan^{- 1} x) \cdot x - \int \frac{1}{1 + x ^{2}} \cdot x d x$
$I = x t a n^{- 1} x - \frac{1}{2} \int \frac{2 x}{1 + x ^{2}} d x$
$I = x t a n^{- 1} x - \frac{1}{2} l o g ∣ 1 + x^{2} ∣ + C$

Q. The integral ∫tan−1xdx is equal to (where, C is the constant of integration)

1+x21​+C

xtan−1x+21​log∣1+x2∣+C

xtan−1x+21​.1+x2tan−1x​+C

xtan−1x−21​log∣1+x2∣+C