Question

Choose the correct answer in ∫√1+x2dx is equal to (A)x2√1+x2+12log∣(x+√1+x2)∣+C(B)23(1+x2)32+C(C)23x(1+x2)32+C(D)x22√1+x2+12x2log∣x+√1+x2∣+C

Answer 1

Answer:

$\frac{1}{2}\ln(x + \sqrt{1+x^2})+\frac{1}{2}x\sqrt{1+x^2}+c$

Step-by-step explanation:

Let x=sinhθ, then set

$I=\int \sqrt{1+x^2}dx$

hence

$\beginI &= \int \sqrt{1+x^2}dx \\&= \int \sqrt{1+\sinh^2 \theta}\cosh \theta d \theta \\&= \int \cosh^2 \theta d \theta \\&= \frac{1}{2}\int 1 +  \cosh 2 \theta d \theta \\&= \frac{1}{2}\theta+\frac{1}{4}\sinh 2 \theta + c\\&= \frac{1}{2}\sinh^{-1}x+\frac{1}{2}x\sqrt{1+x^2}+c\\&=\frac{1}{2}\ln(x + \sqrt{1+x^2})+\frac{1}{2}x\sqrt{1+x^2}+c\end$

please mark it as brainliest