Question

If f(x) = tan^–1 x + ln$\sqrt{1 + x}$ – ln$\sqrt{1 - x}$. The integral of 1/2 f ' (x) with respect to x⁴ is –

• A. $e^{- x^{4}} + c$
• B. – ln (1 – x⁴) + c
• C. $e^{\sqrt{1 - x^{4}}} + c$
• D. ln(1 + x²)+ c

Answer 1

Answer: (B)

– ln (1 – x⁴) + c

Answer 2

f(x) = tan^–1 x + ln$\sqrt{1 + x}$ – ln$\sqrt{1 - x}$

f '(x) = $\frac{1}{1 + x^{2}}$ + $\frac{1}{2(1 + x)}$ + $\frac{1}{2(1 - x)}$

= $\frac { 1 } { 1 + x ^ { 2 } } + \frac { 1 } { 1 - x ^ { 2 } } \Rightarrow \frac { 2 } { \left( 1 - x ^ { 4 } \right) }$

$\frac{1}{2}$ f ' (x) = $\frac{1}{1 - x^{4}}$

– $\int \frac { \mathrm { dt } } { \mathrm { t } }$ ̃ – lnt + c 1 – x⁴ = t

– ln (1 – x⁴) + c – dx⁴ = dt