Question

$\int_{}^{}\cot^{- 1}$ (1 – x + x²) dx = x tan^–1 x – $\frac{1}{2}$log (1 + x²) + A, Then A is equal to –

• A. – (1 – x) tan^–1 (1 – x²) + $\frac{1}{2}$ log {1 + (1 – x)²} + c
• B. (1 – x) tan^–1 (1 – x²) + $\frac{1}{2}$ log {1 + (1 – x)²} + c
• C. – (1 – x) tan^–1 (1 – x) + $\frac{1}{2}$ log {1 + (1 – x)²} + c
• D. None of these

Answer 1

Answer: (C)

– (1 – x) tan^–1 (1 – x) + $\frac{1}{2}$ log {1 + (1 – x)²} + c

Answer 2

We have, $\int_{}^{}\cot^{- 1}$ (1 – x + x²) dx

= $\int \cot ^ { - 1 }${1 – x (1 – x)} dx

= $\int \tan ^ { - 1 }$ $\left\{ \frac{1}{1 - x(1 - x)} \right\}$dx

= $\int \tan ^ { - 1 }$ $\left\{ \frac{x + (1 - x)}{1 - x(1 - x)} \right\}$dx

= $\int \left\{ \tan ^ { - 1 } \right.$ x + tan^–1 (1 – x)} dx

= $\int \tan ^ { - 1 }$x dx + $\int_{}^{}\tan^{- 1}$ (1 – x)dx

= I₁ + I₂ … (1)

where I₁ =$\int_{}^{}\tan^{- 1}$x dx and I₂ = $\int_{}^{}\tan^{- 1}$ (1 – x) dx.

Now, I₁ = $\int_{}^{}\tan^{- 1}$x dx= $\int_{}^{}\tan^{- 1}$x 1 dx

I II

= x tan^–1 x –$\int_{}^{}\frac{x}{1 + x^{2}}$dx

= x tan^–1 x – $\frac { 1 } { 2 }$ $\int_{}^{}\frac{1}{1 + x^{2}}$d (1 + x²)

= x tan^–1 x –$\frac{1}{2}$log (1 + x²)… (2) and

I₂ = $\int_{}^{}\tan^{- 1}$ (1 – x) dx= –$\int_{}^{}\tan^{- 1}$ (1 – x) d (1 – x)

= – $\left\lbrack (1 - x)\tan^{- 1}(1 - x) - \frac{1}{2}\log\{ 1 + (1 - x)^{2}\} \right\rbrack$

[using (2)]

Substituting the values of I₁ and I₂ in (1), we get

$\int_{}^{}\cot^{- 1}$ (1 – x + x²) dx = x tan^–1 x – $\frac{1}{2}$ log (1 + x²)

– (1 – x) tan^–1 (1 – x) + $\frac{1}{2}$ log {1 + (1 – x)²} + c

Hence (3) is the correct answer.