Generating...                               ibee2023_n8

  1. Find $\displaystyle \int {e^{x}\,\cos x\,\sin x}{\;dx}$ by using integration by parts.

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

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

    $\displaystyle \left(-x^2-2\,x-2\right)\,e^ {- x } + C$ $\displaystyle \left(x^2-2\,x+2\right)\,e^{x} + C$ $\displaystyle {{e^{x}\,\left(\sin x-\cos x\right)}\over{2}} + C$ $\displaystyle \left(-x-1\right)\,e^ {- x } + C$ $\displaystyle {{e^{x}\,\left(\sin x+\cos x\right)}\over{2}} + C$ $\displaystyle \left(x-1\right)\,e^{x} + C$

  3. Find $\displaystyle \int {x^2\,\cos x^3}{\;dx}$ by using substitution.

    $\displaystyle {{\cos x^3}\over{3}} + C$ $\displaystyle -{{\sin x^3}\over{3}} + C$ $\displaystyle -{{\cos x^3}\over{3}} + C$ $\displaystyle {{\sin x^3}\over{3}} + C$

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

    $\displaystyle {{2\,x^{{{7}\over{2}}}}\over{7}} + C$ $\displaystyle {{2\,\left(x+1\right)}\over{\sqrt{x}}}-{{\left(x+1\right)^2}\over{2 \,x^{{{3}\over{2}}}}} + C$ $\displaystyle {{2\,x^{{{5}\over{2}}}}\over{5}}+{{4\,x^{{{3}\over{2}}}}\over{3}}+2 \,\sqrt{x} + C$ $\displaystyle {{2\,x^{{{3}\over{2}}}}\over{3}} + C$

  5. Find $\displaystyle \int {\sin ^4x}{\;dx}$ .

    $\displaystyle {{\cos ^3x}\over{3}}-\cos x + C$ $\displaystyle {{\sin \left(2\,x\right)}\over{4}}+{{x}\over{2}} + C$ $\displaystyle {{\sin \left(4\,x\right)}\over{32}}+{{\sin \left(2\,x\right)}\over{ 4}}+{{3\,x}\over{8}} + C$ $\displaystyle {{x}\over{2}}-{{\sin \left(2\,x\right)}\over{4}} + C$ $\displaystyle {{\sin \left(4\,x\right)}\over{32}}-{{\sin \left(2\,x\right)}\over{ 4}}+{{3\,x}\over{8}} + C$ $\displaystyle \sin x-{{\sin ^3x}\over{3}} + C$

  6. Find $\displaystyle \int {\sqrt{\sec x}\,\tan x}{\;dx}$ by using substitution.

    $\displaystyle {{2}\over{\sqrt{\sec x}}} + C$ $\displaystyle 2\,\sqrt{\sec x} + C$ $\displaystyle -{{2}\over{\sqrt{\sec x}}} + C$ $\displaystyle {{2}\over{3\,\left(\sec x\right)^{{{3}\over{2}}}}} + C$

  7. Find $\displaystyle \int {{{\left(\ln x\right)^2}\over{x}}}{\;dx}$ by substituting $\displaystyle u=\ln x$

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

  8. Find $\displaystyle \int {e^{x}\,\left(e^{x}+1\right)^2}{\;dx}$ by substituting $\displaystyle u=e^{x}+1$

    $\displaystyle \int {u^2}{\;du} = {{\left(e^{x}+1\right)^3}\over{3}} + C$ $\displaystyle -\int {u^2}{\;du} = -{{\left(e^{x}+1\right)^3}\over{3}} + C$ $\displaystyle \int {u}{\;du} = {{\left(e^{x}+1\right)^2}\over{2}} + C$ $\displaystyle \int {\left(u-1\right)\,u^2}{\;du} = {{\left(e^{x}+1\right)^4}\over{4}}-{{\left(e^{x}+1\right)^3}\over{3 }} + C$

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

    $\displaystyle {{2\,\left(x+1\right)^{{{3}\over{2}}}}\over{3}}-2\,\sqrt{x+1} + C$ $\displaystyle {{2\,\left(x+1\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle 2\,\sqrt{x+1} + C$ $\displaystyle {{2\,\left(x+1\right)^{{{5}\over{2}}}}\over{5}}-{{2\,\left(x+1 \right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{2\,\left(x+1\right)^{{{7}\over{2}}}}\over{7}}-{{4\,\left(x+1 \... ...{{{5}\over{2}}}}\over{5}}+{{2\,\left(x+1\right)^{{{3}\over{2 }}}}\over{3}} + C$ $\displaystyle {{2\,\left(x+1\right)^{{{5}\over{2}}}}\over{5}}-{{4\,\left(x+1 \right)^{{{3}\over{2}}}}\over{3}}+2\,\sqrt{x+1} + C$

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

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

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

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

  12. Find $\displaystyle \int {{{3\,x-5}\over{x^2-3\,x+2}}}{\;dx}$ .

    $\displaystyle 2\,\ln \left(x+1\right)+\ln \left(x-1\right) + C$ $\displaystyle 2\,\ln \left(x+2\right)+\ln \left(x-1\right) + C$ $\displaystyle \ln \left(x+2\right)+2\,\ln \left(x-1\right) + C$ $\displaystyle 2\,\ln \left(x-1\right)+\ln \left(x-2\right) + C$ $\displaystyle \ln \left(x+1\right)+2\,\ln \left(x-1\right) + C$ $\displaystyle \ln \left(x-1\right)+2\,\ln \left(x-2\right) + C$

  13. Find $\displaystyle \int {\sqrt{x-1}\,x^2}{\;dx}$ by substituting u = x − 1

    $\displaystyle \int {\left(u-1\right)^2\,\sqrt{u}}{\;du} = {{2\,\left(x+1\ri... ...{{{5}\over{2}}}}\over{5}}+{{2\,\left(x+1\right)^{{{3}\over{2 }}}}\over{3}} + C$ $\displaystyle \int {\sqrt{u}\,\left(u+1\right)}{\;du} = {{2\,\left(x-1\right)^{{{5}\over{2}}}}\over{5}}+{{2\,\left(x-1 \right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle \int {\left(u-1\right)\,\sqrt{u}}{\;du} = {{2\,\left(x+1\right)^{{{5}\over{2}}}}\over{5}}-{{2\,\left(x+1 \right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle \int {\sqrt{u}\,\left(u+1\right)^2}{\;du} = {{2\,\left(x-1\ri... ...{{{5}\over{2}}}}\over{5}}+{{2\,\left(x-1\right)^{{{3}\over{2 }}}}\over{3}} + C$

  14. Find $\displaystyle \int {{{x}\over{\sqrt{x^2+1}}}}{\;dx}$ by substituting $\displaystyle x=\sinh \theta$

    $\displaystyle \int {\sinh ^2\theta}{\;d\theta} = {{x\,\sqrt{x^2+1}}\over{2}}-{{{\rm asinh}\; x}\over{2}} + C$ $\displaystyle \int {\cosh ^2\theta\,\sinh ^2\theta}{\;d\theta} = -{{{\rm as... ...,\left(x^2+1\right)^{{{3}\over{2}} }}\over{4}}-{{x\,\sqrt{x^2+1}}\over{8}} + C$ $\displaystyle \int {{{\cosh ^2\theta}\over{\sinh ^2\theta}}}{\;d\theta} = {\rm asinh}\; x-{{\sqrt{x^2+1}}\over{x}} + C$ $\displaystyle \int {{{1}\over{\sinh ^2\theta}}}{\;d\theta} = -{{\sqrt{x^2+1}}\over{x}} + C$ $\displaystyle \int {\sinh \theta}{\;d\theta} = \sqrt{x^2+1} + C$

  15. Find $\displaystyle \int {x\,\sqrt{x^2-4}}{\;dx}$ by substituting $\displaystyle x=2\,\sec \theta$

    $\displaystyle 2\,\int {\left(\sec \theta\right)^2}{\;d\theta} = \sqrt{x^2-4} + C$ $\displaystyle 8\,\int {\left(\sec \theta\right)^2\,\tan ^2\theta}{\;d \theta} = {{\left(x^2-4\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{\int {{{1}\over{\sec \theta}}}{\;d\theta}}\over{4}} = {{\sqrt{x^2-4}}\over{4\,x}} + C$ $\displaystyle \int {\sec \theta}{\;d\theta} = \ln \left(2\,\sqrt{x^2-4}+2\,x\right) + C$


Department of Mathematics
Last modified: 2026-09-10