Generating...                               ibee2025_n7

  1. Find $\displaystyle -\int {{{\sqrt{1-x^2}\,\left(\arcsin x+1\right)}\over{x^2-1}}}{\;dx }$

    $\displaystyle 2\,\sqrt{\arcsin x+1} + C$ $\displaystyle \ln \left(\arcsin x+1\right) + C$ $\displaystyle {{\left(\arcsin x+1\right)^2}\over{2}} + C$ $\displaystyle \arcsin x + C$ $\displaystyle {{2\,\left(\arcsin x+1\right)^{{{3}\over{2}}}}\over{3}} + C$

  2. Find $\displaystyle \int {{{1}\over{x^2\,\sqrt{x^2-9}}}}{\;dx}$ by substituting $\displaystyle x=3\,\sec \theta$

    $\displaystyle \int {\sec \theta}{\;d\theta} = \ln \left(2\,\sqrt{x^2-9}+2\,x\right) + C$ $\displaystyle 3\,\int {\left(\sec \theta\right)^2}{\;d\theta} = \sqrt{x^2-9} + C$ $\displaystyle 27\,\int {\left(\sec \theta\right)^2\,\tan ^2\theta}{\;d \theta} = {{\left(x^2-9\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{\int {{{1}\over{\sec \theta}}}{\;d\theta}}\over{9}} = {{\sqrt{x^2-9}}\over{9\,x}} + C$

  3. Find $\displaystyle \int {{{1}\over{\sqrt{9-x^2}}}}{\;dx}$.

    $\displaystyle {{\ln \left(x^2+9\right)}\over{2}} + C$ $\displaystyle {{\ln \left(x+3\right)}\over{6}}-{{\ln \left(x-3\right)}\over{6}} + C$ $\displaystyle \arcsin \left({{x}\over{3}}\right) + C$ $\displaystyle {{\arctan \left({{x}\over{3}}\right)}\over{3}} + C$ $\displaystyle {\rm asinh}\; \left({{x}\over{3}}\right) + C$ $\displaystyle \ln \left(2\,\sqrt{x^2-9}+2\,x\right) + C$

  4. Find $\displaystyle -\int {\cos ^2x\,\sin ^5x}{\;dx}$ by using substitution.

    $\displaystyle {{\cos ^3x}\over{3}} + C$ $\displaystyle {{\cos ^5x}\over{5}} + C$ $\displaystyle {{\cos ^6x}\over{6}}-{{\cos ^4x}\over{2}}+{{\cos ^2x}\over{2}} + C$ $\displaystyle {{\cos ^7x}\over{7}}-{{2\,\cos ^5x}\over{5}}+{{\cos ^3x}\over{3}} + C$

  5. Find $\displaystyle \int {{{x}\over{x^2+3}}}{\;dx}$ by substituting $\displaystyle u=x^2+3$

    $\displaystyle \int {{{1}\over{u^2}}}{\;du} = -{{1}\over{x^2+3}} + C$ $\displaystyle {{\int {{{1}\over{u}}}{\;du}}\over{2}} = {{\ln \left(x^2+3\right)}\over{2}} + C$ $\displaystyle {{\int {{{u-3}\over{u}}}{\;du}}\over{2}} = {{x^2+3}\over{2}}-{{3\,\ln \left(x^2+3\right)}\over{2}} + C$ $\displaystyle \int {{{1}\over{u}}}{\;du} = \ln \left(x^2+3\right) + C$


Department of Mathematics
Last modified: 2025-10-30