Question

${\int }_{-1}^1\frac{3x^{2}dx}{1+4^{tanx}}=$

• A. -1
• B. 1
• C. 2

Answer 1

The correct option is B 1

$I={\int }_{-1}^1\frac{3x^{2}dx}{1+4^{tanx}}$
$I={\int }_{-1}^1\frac{3x^{2}dx{.4}^{tanx}}{4^{tanx}+1}\{\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx\}$
$\therefore 2I={\int }_{-1}^{1}3x^2\Rightarrow I=1$