Generating...                               ibee2025_n24

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

    $\displaystyle {{\ln \left(x^2+9\right)}\over{2}}+\ln x-{{\arctan \left({{x }\over{3}}\right)}\over{3}} + C$ $\displaystyle \ln x+{{\arctan \left({{x}\over{3}}\right)}\over{3}} + C$ $\displaystyle {{\ln \left(x^2+9\right)}\over{2}}+{{\arctan \left({{x}\over{3}} \right)}\over{3}} + C$ $\displaystyle {{\ln \left(x^2+9\right)}\over{2}}+\ln x+{{\arctan \left({{x }\over{3}}\right)}\over{3}} + C$ $\displaystyle \ln x-{{\arctan \left({{x}\over{3}}\right)}\over{3}} + C$

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

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

  3. Find $\displaystyle \int {x^2\,e^{x}}{\;dx}$ by using integration by parts.

    $\displaystyle \left(-x^2-2\,x-2\right)\,e^ {- x } + C$ $\displaystyle {{2\,e^{x}\,\sin \left(2\,x\right)+e^{x}\,\cos \left(2\,x\right)+5 \,e^{x}}\over{10}} + C$ $\displaystyle \left(x^2-2\,x+2\right)\,e^{x} + C$ $\displaystyle -{{2\,e^{x}\,\sin \left(2\,x\right)+e^{x}\,\cos \left(2\,x\right)-5 \,e^{x}}\over{10}} + C$ $\displaystyle {{e^{x}\,\left(\sin \left(2\,x\right)-2\,\cos \left(2\,x\right) \right)}\over{10}} + C$

  4. Find $\displaystyle \int {{{x}\over{\sqrt{x^2+4}}}}{\;dx}$ by substituting $\displaystyle x=2\,\sinh \theta$

    $\displaystyle 16\,\int {\cosh ^2\theta\,\sinh ^2\theta}{\;d\theta} = {{x\,\... ...}-{{x\,\sqrt{x^2+4} }\over{2}}-2\,{\rm asinh}\; \left({{x}\over{2}}\right) + C$ $\displaystyle {{\int {{{1}\over{\sinh ^2\theta}}}{\;d\theta}}\over{4}} = -{{\sqrt{x^2+4}}\over{4\,x}} + C$ $\displaystyle 2\,\int {\sinh \theta}{\;d\theta} = \sqrt{x^2+4} + C$ $\displaystyle \int {{{\cosh ^2\theta}\over{\sinh ^2\theta}}}{\;d\theta} = {\rm asinh}\; \left({{x}\over{2}}\right)-{{\sqrt{x^2+4}}\over{x}} + C$ $\displaystyle 4\,\int {\sinh ^2\theta}{\;d\theta} = {{x\,\sqrt{x^2+4}}\over{2}}-2\,{\rm asinh}\; \left({{x}\over{2}} \right) + C$

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

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


Department of Mathematics
Last modified: 2026-09-10