Generating...                               ibee2025_n29

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

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

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

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

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

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

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

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

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

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


Department of Mathematics
Last modified: 2025-05-03