Generating...                               ibee2023_n17

  1. Find $\displaystyle \int {\left(\sec x\right)^4\,\tan x}{\;dx}$ by substituting $\displaystyle u=\tan x$

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

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

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

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

    $\displaystyle {{\int {{{1}\over{\sinh ^2\theta}}}{\;d\theta}}\over{9}} = -{{\sqrt{x^2+9}}\over{9\,x}} + C$ $\displaystyle 27\,\int {\cosh ^2\theta\,\sinh \theta}{\;d\theta} = {{\left(x^2+9\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle 81\,\int {\cosh ^2\theta\,\sinh ^2\theta}{\;d\theta} = {{x\,\... ...2 +9}}\over{8}}-{{81\,{\rm asinh}\; \left({{x}\over{3}}\right)}\over{8 }} + C$ $\displaystyle \int {{{\cosh ^2\theta}\over{\sinh ^2\theta}}}{\;d\theta} = {\rm asinh}\; \left({{x}\over{3}}\right)-{{\sqrt{x^2+9}}\over{x}} + C$ $\displaystyle 3\,\int {\sinh \theta}{\;d\theta} = \sqrt{x^2+9} + C$

  4. Find $\displaystyle \int {\sec x}{\;dx}$ .

    $\displaystyle \ln \sec x + C$ $\displaystyle \tan x-x + C$ $\displaystyle \ln \left(\tan x+\sec x\right) + C$ $\displaystyle \tan x + C$

  5. Find $\displaystyle \int {\cos x\,\sin ^5x}{\;dx}$ by substituting $\displaystyle u=\sin x$

    $\displaystyle \int {u^5\,\sqrt{1-u^2}}{\;du} = -{{\sin ^4x\,\left(1-\sin ^2... ...{2}}}}\over{35}}-{{8\, \left(1-\sin ^2x\right)^{{{3}\over{2}}}}\over{105}} + C$ $\displaystyle \int {u^5}{\;du} = {{\sin ^6x}\over{6}} + C$ $\displaystyle \int {u^4}{\;du} = {{\sin ^5x}\over{5}} + C$ $\displaystyle -\int {u^5}{\;du} = -{{\sin ^6x}\over{6}} + C$

  6. Find $\displaystyle \int {{{x^2}\over{x^2+1}}}{\;dx}$.

    $\displaystyle {{2\,x}\over{\left(x^2+1\right)^2}} + C$ $\displaystyle x+{{1}\over{x}} + C$ $\displaystyle {{\ln \left(x+1\right)}\over{2}}+x-{{\ln \left(x-1\right)}\over{2 }} + C$ $\displaystyle x-\arctan x + C$

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

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

  8. Find $\displaystyle \int {{{1}\over{x^2+4}}}{\;dx}$.

    $\displaystyle {{\arctan \left({{x}\over{2}}\right)}\over{2}} + C$ $\displaystyle {{\ln \left(x^2+4\right)}\over{2}} + C$ $\displaystyle {{\ln \left(x+2\right)}\over{4}}-{{\ln \left(x-2\right)}\over{4}} + C$ $\displaystyle {\rm asinh}\; \left({{x}\over{2}}\right) + C$ $\displaystyle \ln \left(2\,\sqrt{x^2-4}+2\,x\right) + C$ $\displaystyle \arcsin \left({{x}\over{2}}\right) + C$

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

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

  10. Find $\displaystyle \int {{{1}\over{2\,x+5}}}{\;dx}$ by using substitution.

    $\displaystyle -{{\ln \left(2\,x+5\right)}\over{2}} + C$ $\displaystyle -{{1}\over{2\,x+5}} + C$ $\displaystyle -\ln \left(2\,x+5\right) + C$ $\displaystyle {{\ln \left(2\,x+5\right)}\over{2}} + C$ $\displaystyle \ln \left(2\,x+5\right) + C$

  11. Find $\displaystyle \int {\cos ^2x}{\;dx}$ .

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

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

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

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

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

  14. Find $\displaystyle \int {x\,\sqrt{x+2}}{\;dx}$.

    $\displaystyle {{2\,\left(x+2\right)^{{{3}\over{2}}}}\over{3}}-4\,\sqrt{x+2} + C$ $\displaystyle 2\,\sqrt{x+2} + C$ $\displaystyle {{2\,\left(x+2\right)^{{{5}\over{2}}}}\over{5}}-{{4\,\left(x+2 \right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{2\,\left(x+2\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{2\,\left(x+2\right)^{{{7}\over{2}}}}\over{7}}-{{8\,\left(x+2 \... ...{{{5}\over{2}}}}\over{5}}+{{8\,\left(x+2\right)^{{{3}\over{2 }}}}\over{3}} + C$ $\displaystyle {{2\,\left(x+2\right)^{{{5}\over{2}}}}\over{5}}-{{8\,\left(x+2 \right)^{{{3}\over{2}}}}\over{3}}+8\,\sqrt{x+2} + C$

  15. Find $\displaystyle \int {\left(\sec x\right)^2\,\tan ^5x}{\;dx}$ by substituting $\displaystyle u=\sec x$

    $\displaystyle \int {u}{\;du} = {{\left(\sec x\right)^2}\over{2}} + C$ $\displaystyle \int {u^3}{\;du} = {{\left(\sec x\right)^4}\over{4}} + C$ $\displaystyle \int {\left(u^2-1\right)^2}{\;du} = {{\left(\sec x\right)^5}\over{5}}-{{2\,\left(\sec x\right)^3}\over{ 3}}+\sec x + C$ $\displaystyle \int {u\,\left(u^2-1\right)^2}{\;du} = {{\left(\sec x\right)^... ...r{6}}-{{\left(\sec x\right)^4}\over{2}} +{{\left(\sec x\right)^2}\over{2}} + C$


Department of Mathematics
Last modified: 2026-10-30