Category: Integration

Question Number 55067 by gunawan last updated on 17/Feb/19 $$\mathrm{Known}C\mathrm{circle}\mathrm{centered}\mathrm{at}0.$$$$\mathrm{Find}\mathrm{value}\mathrm{than}{\int }_C\frac{dz}{1-z}$$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 120599 by MagdiRagheb last updated on 01/Nov/20 $$Letx=u^{6}dx=6u^{5}du$$$$I=\int \frac{u^3}{{(1+u^2)}^2}\times 6u^{5}du$$$$=6\int \frac{u^8}{1+2u^2+u^4}du$$$$\:\:\:=\mathrm{6}\:\int\left[\frac{−\mathrm{4}}{{u}^{\mathrm{2}}…

Question Number 120582 by MagdiRagheb last updated on 01/Nov/20 $$Letu=x^{\frac{3}{5}}du=\frac{3}{5x^{\frac{2}{5}}}$$$$I=\frac{5}{3}\int \frac{u}{\sqrt{3-2u}}du=-\frac{5}{6}\int \frac{3-2u-3}{\sqrt{3-2u}}du$$$$=-\frac{5}{3}\int [\sqrt{3-2u}-3{(3-2u)}^{-\frac{1}{2}}]du$$$$=-\frac{5}{3}[-\frac{1}{3}{(3-2u)}^{\frac{3}{2}}+3{(3-2u)}^{\frac{1}{2}}]+c$$$$=\:−\frac{\mathrm{5}}{\mathrm{18}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left[−\mathrm{3}+\mathrm{2}{u}\:+\:\mathrm{9}\right]+{c} \…

Question Number 120562 by peter frank last updated on 01/Nov/20 Answered by Dwaipayan Shikari last updated on 01/Nov/20 $$\int 1-\frac{bx+12}{x^2+bx+12}dx$$$$=x-b\int \frac{x+\frac{12}{b}}{x^2+bx+12}dx$$$$={x}−\frac{{b}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+{b}}{{x}^{\mathrm{2}}…

Question Number 120552 by talminator2856791 last updated on 01/Nov/20 $$$$$$$$$$\mathrm{evaluate}:{\int }_0^{\mathrm{\infty }}{\left(\frac{x}{x+1}\right)}^{x!}dx$$$$$$$$$$ Terms of…