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Question Number 32302 by abdo imad last updated on 22/Mar/18

$$calculate{\int }_1^2\frac{dx}{x+x\sqrt{x}}.$$

Answered by sma3l2996 last updated on 23/Mar/18

$$lett=\sqrt{x}\Rightarrow dx=2tdt$$$$A={\int }_1^2\frac{dx}{x+x\sqrt{x}}=2{\int }_1^{\sqrt{2}}\frac{tdt}{t^2(1+t)}=2{\int }_1^{\sqrt{2}}\frac{dt}{t(1+t)}=2{\int }_1^{\sqrt{2}}(\frac{1}{t}-\frac{1}{t+1})dt$$$$=2{[ln\mid \frac{t}{t+1}\mid ]}_1^{\sqrt{2}}=2(ln\left(\frac{\sqrt{2}}{\sqrt{2}+1}\right)+ln2)$$

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