Answer to Question #162040 in Calculus for Phyroe

Question

Evaluate the integral of √x dx/1+x√x

Answer 1

$\int\frac{\sqrt xdx}{1+x\sqrt x}$

Solution:

$I=\int\frac{\sqrt xdx}{1+x\sqrt x}$

$\sqrt x=t;$ $x=t^2;$ $dx=2tdt$ .

$I=\int \frac{2t^2dt}{1+t^3}$

$\frac{2t^2dt}{1+t^3}=\frac{2t^2}{(t+1)(t^2-t+1)}=\frac{2t^2}{(t+1)((t-\frac12)^2+\frac34)}$

Expand the fraction:

$\frac{2t^2}{(t+1)((t-\frac12)^2+\frac34)}=\frac{2}{3(t+1)}+\frac{\frac43(t-\frac12)}{(t-\frac12)^2+\frac34}$

$I=\int (\frac{2}{3(t+1)}+\frac{\frac43(t-\frac12)}{(t-\frac12)^2+\frac34})dt=$

$\int \frac{2dt}{3(t+1)}+\int\frac{\frac23d(t-\frac12)^2}{(t-\frac12)^2+\frac34}=$

$\frac23(\ln{|t+1|}+\ln{{((t-\frac12)^2+\frac34})})+C=$

$\frac23\ln{|1+t^3|}+C=\frac23\ln{|1+x\sqrt x|}+C=\ln{(1+x\sqrt x)^{\frac23}}+C$

Answer: $\int\frac{\sqrt xdx}{1+x\sqrt x}=\ln{(1+x\sqrt x)^{\frac23}}+C$