Generating...                               s20quiz18_n13

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

    $\displaystyle 2\,\sqrt{x} + C$ $\displaystyle {{2\,x^{{{3}\over{2}}}}\over{3}}+2\,\sqrt{x} + C$ $\displaystyle {{1}\over{\sqrt{x}}}-{{x+1}\over{2\,x^{{{3}\over{2}}}}} + C$ $\displaystyle {{2\,x^{{{5}\over{2}}}}\over{5}} + C$

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

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

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

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

  4. Find $\displaystyle \int {x^3\,\left(x^4+2\right)^3}{\;dx}$ by substituting $\displaystyle u=x^4+2$

    $\displaystyle \int {u^3}{\;du} = {{\left(x^4+2\right)^4}\over{4}} + C$ $\displaystyle {{\int {u^3}{\;du}}\over{4}} = {{\left(x^4+2\right)^4}\over{16}} + C$ $\displaystyle \int {u^2}{\;du} = {{\left(x^4+2\right)^3}\over{3}} + C$ $\displaystyle {{\int {\left(u-2\right)\,u^3}{\;du}}\over{4}} = {{\left(x^4+2\right)^5}\over{20}}-{{\left(x^4+2\right)^4}\over{8}} + C$

  5. Evaluate $\displaystyle \int_{-3}^{-2}{e^{x+3}\;dx}$.

    $\displaystyle\Big[ e^{x+3} \Big]_{ -3}^{ -2} = e-1$ $\displaystyle\Big[ e^{x} \Big]_{ -3}^{ -2} = e^ {- 2 }-e^ {- 3 }$ $\displaystyle\Big[ e^{x+4} \Big]_{ -3}^{ -2} = e^2-e$ $\displaystyle\Big[ 3\,e^{x} \Big]_{ -3}^{ -2} = 3\,e^ {- 2 }-3\,e^ {- 3 }$

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

    $\displaystyle {{x^2}\over{2}}+x + C$ $\displaystyle 2\,\sqrt{x} + C$ $\displaystyle {{\left(x+1\right)^2}\over{x}} + C$ $\displaystyle {{2\,x^{{{3}\over{2}}}+6\,\sqrt{x}}\over{3}} + C$

  7. Evaluate $\displaystyle \int_{{{\pi}\over{4}}}^{{{\pi}\over{2}}}{{{\sin ^2x+1}\over{\sin ^2 x}}\;dx}$.

    $\displaystyle\Big[ -\cot x \Big]_{ {{\pi}\over{4}}}^{ {{\pi}\over{2}}} = 1$ $\displaystyle\Big[ {{1}\over{\sin ^2x}}+1 \Big]_{ {{\pi}\over{4}}}^{ {{\pi}\over{2}}} = -1$ $\displaystyle\Big[ x-\cos x \Big]_{ {{\pi}\over{4}}}^{ {{\pi}\over{2}}} = {{\sqrt{2}\,\pi+4}\over{2^{{{5}\over{2}}}}}$ $\displaystyle\Big[ x-\cot x \Big]_{ {{\pi}\over{4}}}^{ {{\pi}\over{2}}} = {{\pi+4}\over{4}}$

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

    $\displaystyle -\cot x-x + C$ $\displaystyle -\cot x + C$ $\displaystyle \cot x + C$ $\displaystyle x-\cot x + C$

  9. Evaluate $\displaystyle \int_{1}^{4}{{{1}\over{\sqrt{x}}}\;dx}$.

    $\displaystyle\Big[ {{1}\over{x}} \Big]_1^4 = -{{3}\over{4}}$ $\displaystyle\Big[ 2\,\sqrt{x} \Big]_1^4 = 2$ $\displaystyle\Big[ x \Big]_1^4 = 3$ $\displaystyle\Big[ 2\,\sqrt{x} \Big]_1^4 = 2$

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

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


Department of Mathematics
Last modified: 2026-07-16