Question

The anti derivative of $\int \biggr\lgroup{\sqrt{x}+\frac{1}{\sqrt{x}}\biggr\rgroup \, dx$ equals,
(A)$\frac 13 x^{\frac 13}+ 2x^{\frac 12} +C$
(B) $\frac 23 x^{\frac 23}+ \frac 12 x^{2} +C$
(C) $\frac 23 x^{\frac 32}+ 2x^{\frac 12} +C$
(D) $\frac 32 x^{\frac 32}+ \frac 12 x^{\frac 12} + C$

Answer 1

Answer:

option (c) is correct

Step-by-step explanation:

Formula used:

$\int{x^n}\:dx=\frac{x^{n+1}}{n+1}+c$

Now,

$\int[\sqrt{x}+\frac{1}{\sqrt{x}}]\:dx\\\\=\int[x^{\frac{1}{2}}+x^\frac{-1}{2}}]\:dx\\\\=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+c\\\\=\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+c\\\\=\frac{2}{3}x\sqrt{x}+2\sqrt{x}+c$