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Question Number 118230 by bobhans last updated on 16/Oct/20

$$\mathrm{If}\int \frac{{\left(\sqrt{\mathrm{x}}\right)}^5}{{\left(\sqrt{\mathrm{x}}\right)}^7+{\mathrm{x}}^6}\mathrm{dx}=\mathrm{p}\ln \left(\frac{{\mathrm{x}}^{\mathrm{q}}}{{\mathrm{x}}^{\mathrm{q}}+1}\right)+\mathrm{C}$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{p}\mathrm{and}\mathrm{q}.$$

Answered by bemath last updated on 16/Oct/20

$$I=\int \frac{x^2\sqrt{x}}{x^3\sqrt{x}+x^6}dx=\int \frac{\sqrt{x}}{x\sqrt{x}+x^4}dx$$$$I=\int \frac{dx}{x+x^3\sqrt{x}}.$$$$let\sqrt{x}=u\to dx=2\sqrt{x}du$$$$I=\int \frac{2udu}{u^2+u^7}=\int \frac{2du}{u+u^6}$$$$I=\int \frac{2du}{u(1+u^5)}=\int \frac{2(1+u^5-u^5)}{u(1+u^5)}du$$$$=2\int (\frac{1}{u}-\frac{u^4}{1+u^5})du$$$$=2(\ln u-\frac{1}{5}\int \frac{d(1+u^5)}{1+u^5})$$$$=2(\ln u-\frac{1}{5}\ln (1+u^5))+c$$$$=\frac{2}{5}(5\ln u-\ln (1+u^5))+c$$$$=\frac{2}{5}\ln \left(\frac{u^5}{1+u^5}\right)+c=p\ln \left(\frac{x^q}{1+x^q}\right)+c$$$$=\frac{2}{5}\ln \left(\frac{x^{\frac{5}{2}}}{1+x^{\frac{5}{2}}}\right)+c=p\ln \left(\frac{x^q}{1+x^q}\right)+c$$$$weget\{\begin{array}{l}p=\frac{2}{5}\\ q=\frac{5}{2}\end{array}$$

Commented by bobhans last updated on 16/Oct/20

$$correct!$$

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