Question

$\int_{}^{}{\mathbf{x}\sqrt{\mathbf{1 +}\mathbf{x}^{\mathbf{2}}}\mathbf{dx =}}$

• A. $\frac{1 + 2x^{2}}{\sqrt{1 + x^{2}}} + c$
• B. $\sqrt{1 + x^{2}} + c$
• C. $3(1 + x^{2})^{3/2} + c$
• D. $\frac{1}{3}(1 + x^{2})^{3/2} + c$

Answer 1

Answer: (D)

$\frac{1}{3}(1 + x^{2})^{3/2} + c$

Answer 2

Put$1 + x^{2} = t$ $\Rightarrow 2xdx = dt$,

$\int_{}^{}{x\sqrt{1 + x^{2}}dx = \frac{1}{2}\int_{}^{}{t^{1/2}dt = \frac{1}{2}.\frac{t^{3/2}}{3/2} + c}}$ $= \frac{1}{3}(1 + x^{2})^{3/2} + c$