Generating...                               ibee2025_n1

  1. Find $\displaystyle \int {{{1}\over{x\,\sqrt{\ln x}}}}{\;dx}$ by substituting $\displaystyle u=\ln x$

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

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

    $\displaystyle x\,\ln x-x + C$ $\displaystyle 2\,\sqrt{x}\,\ln x-4\,\sqrt{x} + C$ $\displaystyle {{x^3\,\ln x}\over{3}}-{{x^3}\over{9}} + C$ $\displaystyle {{x^2\,\ln x}\over{2}}-{{x^2}\over{4}} + C$ $\displaystyle -{{\ln x}\over{x}}-{{1}\over{x}} + C$ $\displaystyle {{2\,x^{{{3}\over{2}}}\,\ln x}\over{3}}-{{4\,x^{{{3}\over{2}}} }\over{9}} + C$

  3. Find $\displaystyle \int {e^{x}\,\sqrt{e^{x}+1}}{\;dx}$ by substituting $\displaystyle u=e^{x}+1$

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

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

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

  5. Find $\displaystyle \int {{{\cos \sqrt{x}}\over{\sqrt{x}}}}{\;dx}$ by using substitution.

    $\displaystyle 2\,\cos \sqrt{x} + C$ $\displaystyle -2\,\sin \sqrt{x} + C$ $\displaystyle -2\,\cos \sqrt{x} + C$ $\displaystyle 2\,\sin \sqrt{x} + C$


Department of Mathematics
Last modified: 2026-02-06