Question #97142
Evaluate
∫√(1+√x) dx
1
Expert's answer
2019-10-27T12:40:51-0400

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


ddx(xn)=n×xn1ddx(constant)=0\frac {d} {dx} (x^n) = n \times x^ {n-1}\\ \frac {d}{dx} (constant ) =0


Let 1+x=tLet \space 1 + \sqrt x = t

Differentiate with respect to x

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

0+12xdx=dt0 + \frac {1} {2 \sqrt x} dx = dt

12xdx=dt\frac {1} {2 \sqrt x} dx = dt

dx=2xdtdx = 2 \sqrt x dt

dx=2(t1)dtdx = 2 ( t -1 ) dt

(1+x)dx=(t)2(t1)dt\int \sqrt {( 1 + \sqrt x)} dx = \int (\sqrt t) 2 (t -1) dt

=2(t32t12)dt+c=2 \int ( t^ {\frac {3}{2} } -t^ {\frac {1}{2}} ) dt +c

=2(t32+132+1t12+112+1)+c= 2 ( \frac {t^ {\frac {3}{2} +1}}{\frac {3}{2} +1} - \frac {t^ {\frac {1}{2} +1}}{\frac {1}{2} +1} ) +c


=2(t5252t3232)+c= 2 ( \frac {t^ {\frac {5}{2} }}{\frac {5}{2}} - \frac {t^ {\frac {3}{2} }}{\frac {3}{2} } ) +c

=4(t525t323)+c= 4 ( \frac {t^ {\frac {5}{2} }}{5} - \frac {t^ {\frac {3}{2} }}{3} ) +c




=4((1+x)525(1+x)323)+c= 4 ( \frac {(1+ \sqrt x)^ {\frac {5}{2} }}{5} - \frac {(1 + \sqrt x)^ {\frac {3}{2} }}{3} ) +c

Answer :

(1+x)dx=4((1+x)525(1+x)323)+c\int \sqrt {( 1 + \sqrt x)} dx = 4 ( \frac {(1+ \sqrt x)^ {\frac {5}{2} }}{5} - \frac {(1 + \sqrt x)^ {\frac {3}{2} }}{3} ) +c


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