Answer to Question #97142 in Calculus for Anand

Question

Evaluate
∫√(1+√x) dx

Answer 1

Solution:

We need to find the Integration of given function.

We can use the suitable substitution to change to simple form

We have formulae about differentiation

$$\frac {d} {dx} (x^n) = n \times x^ {n-1}\\ \frac {d}{dx} (constant ) =0$$

$$Let \space 1 + \sqrt x = t$$

Differentiate with respect to x

$$d(1 + \sqrt x) = dt$$

$$0 + \frac {1} {2 \sqrt x} dx = dt$$

$$\frac {1} {2 \sqrt x} dx = dt$$

$$dx = 2 \sqrt x dt$$

$$dx = 2 ( t -1 ) dt$$

$$\int \sqrt {( 1 + \sqrt x)} dx = \int (\sqrt t) 2 (t -1) dt$$

$$=2 \int ( t^ {\frac {3}{2} } -t^ {\frac {1}{2}} ) dt +c$$

$$= 2 ( \frac {t^ {\frac {3}{2} +1}}{\frac {3}{2} +1} - \frac {t^ {\frac {1}{2} +1}}{\frac {1}{2} +1} ) +c$$

$$= 2 ( \frac {t^ {\frac {5}{2} }}{\frac {5}{2}} - \frac {t^ {\frac {3}{2} }}{\frac {3}{2} } ) +c$$

$$= 4 ( \frac {t^ {\frac {5}{2} }}{5} - \frac {t^ {\frac {3}{2} }}{3} ) +c$$

$$= 4 ( \frac {(1+ \sqrt x)^ {\frac {5}{2} }}{5} - \frac {(1 + \sqrt x)^ {\frac {3}{2} }}{3} ) +c$$

Answer :

$$\int \sqrt {( 1 + \sqrt x)} dx = 4 ( \frac {(1+ \sqrt x)^ {\frac {5}{2} }}{5} - \frac {(1 + \sqrt x)^ {\frac {3}{2} }}{3} ) +c$$