Question

Solve $\int \frac{x}{{(7x-10-x^2)}^{3/2}}dx$

Answer 1

$\int \frac{x}{{(7x-10-x^2)}^{3/2}}dx=\frac{-1}{2}\int \frac{-2x}{{(7x-10-x^2)}^{3/2}}dx$

$=\frac{-1}{2}[\int \frac{(7-2x)}{{(7x-10-x^2)}^{3/2}}dx-\int \frac{7}{{(7x-10-x^2)}^{3/2}}dx]$

$A=\int \frac{(7-2x)}{{(7x-10-x^2)}^{3/2}}dx$, Let $7x-10-x^2=m;dm=(7-2x)dx$

$=\int \frac{dm}{m^{3/2}}$

$=\frac{m^{-3/2+1}}{-3/2+1}+C=\frac{-2}{\sqrt{m}}+C=\frac{-2}{{(7x-10-x^2)}^{1/2}}+C$

$B=\int \frac{7}{{(7x-10-x^2)}^{3/2}}dx$$=7\int \frac{1}{{(\frac{9}{4}-{(x-\frac{7}{2})}^2)}^{3/2}}dx$; Let $x-\frac{7}{2}=\frac{3}{2}y;\therefore dx=\frac{3}{2}dy$

$=7\int \frac{1}{{(\frac{9}{4}-\frac{9}{4}{y}^2)}^{3/2}}\left(\frac{3}{2}\right)dy$

$=7\int \frac{\left(3/2\right)}{{\left(\frac{3}{2}\right)}^3{(1-y^2)}^{3/2}}dy$

$=\frac{28}{9}\int \frac{1}{{(1-y^2)}^{3/2}}dy$ Let $y=\cos \theta$

$dy=-\sin \theta d\theta$

$=\frac{28}{9}\int \frac{1(-\sin \theta )}{{(1-{\cos }^2\theta )}^{3/2}}d\theta$

$=\frac{28}{9}\int \frac{-\sin \theta }{{\sin }^3\theta }d\theta$

$=\frac{28}{9}\int -{\csc }^2\theta d\theta$

$=\frac{28}{9}\cot \theta +k$

$=\frac{28}{9}\left(\frac{y}{\sqrt{1-y^2}}\right)+k$

$=\frac{28}{9}\left(\frac{\frac{2}{3}(x-\frac{7}{2})}{{(1-\frac{4}{9}{(x-\frac{7}{2})}^2)}^{1/2}}\right)+k$

$=\frac{28}{27}\left(\frac{(2x-7)}{{\left(\frac{4}{9}\right)}^{1/2}{(\frac{9}{4}-{(x-\frac{7}{2})}^2)}^{1/2}}\right)+k$

$=\frac{14}{9}\left(\frac{2x-7}{{(7x-10-x^2)}^{1/2}}\right)+k$

$\therefore \int \frac{x}{{(7x-10-x^2)}^{3/2}}dx=\frac{-1}{2}(A-B)$

$=\frac{-1}{2}\left(\frac{-2-\frac{28x}{9}+\frac{98}{9}}{{(7x-10-x^2)}^{1/2}}\right)+k_1$

$=\frac{2}{9}\left(\frac{7x-40}{{(7x-10-x^2)}^{1/2}}\right)+k_1$