Evaluate $\displaystyle \int_{1}^{4}{{{x+1}\over{\sqrt{x}}}\;dx}$ .
$\displaystyle\Big[ {{2\,x^{{{3}\over{2}}}}\over{3}}+2\,\sqrt{x} \Big]_1^4 = {{20}\over{3}}$ $\displaystyle\Big[ 2\,\sqrt{x} \Big]_1^4 = 2$ $\displaystyle\Big[ {{x^2}\over{2}}+x \Big]_1^4 = {{21}\over{2}}$ $\displaystyle\Big[ {{\left(x+1\right)^2}\over{x}} \Big]_1^4 = {{9}\over{4}}$
Find $\displaystyle \int {{{x^3}\over{\sqrt{x^4+2}}}}{\;dx}$ by substituting $\displaystyle u=x^4+2$
$\displaystyle \int {{{1}\over{u^{{{3}\over{2}}}}}}{\;du} = -{{2}\over{\sqrt{x^4+2}}} + C$ $\displaystyle {{\int {{{u-2}\over{\sqrt{u}}}}{\;du}}\over{4}} = {{\left(x^4+2\right)^{{{3}\over{2}}}}\over{6}}-\sqrt{x^4+2} + C$ $\displaystyle {{\int {{{1}\over{\sqrt{u}}}}{\;du}}\over{4}} = {{\sqrt{x^4+2}}\over{2}} + C$ $\displaystyle \int {{{1}\over{\sqrt{u}}}}{\;du} = 2\,\sqrt{x^4+2} + C$
Find $\displaystyle \int {{{x^2}\over{x+1}}}{\;dx}$ by substituting u = x + 1
$\displaystyle \int {{{\left(u+1\right)^2}\over{u}}}{\;du} = 2\,\left(x-1\right)+{{\left(x-1\right)^2}\over{2}}+\ln \left(x-1 \right) + C$ $\displaystyle \int {{{\left(u-1\right)^2}\over{u}}}{\;du} = \ln \left(x+1\right)+{{x^2-2\,x}\over{2}} + C$ $\displaystyle \int {{{u+1}\over{u}}}{\;du} = x+\ln \left(x-1\right)-1 + C$ $\displaystyle \int {{{u-1}\over{u}}}{\;du} = -\ln \left(x+1\right)+x+1 + C$
Evaluate $\displaystyle \int_{0}^{1}{\left(\sqrt{x}+3\right)^2\;dx}$ .
$\displaystyle\Big[ {{x^2}\over{2}}+4\,x^{{{3}\over{2}}}+9\,x \Big]_0^1 = {{27}\over{2}}$ $\displaystyle\Big[ {{2\,x^{{{3}\over{2}}}}\over{3}}+3\,x \Big]_0^1 = {{11}\over{3}}$ $\displaystyle\Big[ {{2\,x^{{{3}\over{2}}}}\over{3}} \Big]_0^1 = {{2}\over{3}}$ $\displaystyle\Big[ x+6\,\sqrt{x}+9 \Big]_0^1 = 7$
Find $\displaystyle \int {{{1}\over{x\,\ln x}}}{\;dx}$ by substituting $\displaystyle u=\ln x$
$\displaystyle \int {e^{u}}{\;du} = x + C$ $\displaystyle \int {{{1}\over{u^2}}}{\;du} = -{{1}\over{\ln x}} + C$ $\displaystyle -\int {u}{\;du} = -{{\left(\ln x\right)^2}\over{2}} + C$ $\displaystyle \int {{{1}\over{u}}}{\;du} = \ln \ln x + C$
Find $\displaystyle \int {{{1-x}\over{\sqrt{x}}}}{\;dx}$ .
$\displaystyle 2\,\sqrt{x} + C$ $\displaystyle x-{{x^2}\over{2}} + C$ $\displaystyle {{\left(1-x\right)^2}\over{x}} + C$ $\displaystyle -{{2\,x^{{{3}\over{2}}}-6\,\sqrt{x}}\over{3}} + C$
Find $\displaystyle \int {x^2\,\sin x^3}{\;dx}$ by using substitution.
$\displaystyle {{\sin x^3}\over{3}} + C$ $\displaystyle -{{\sin x^3}\over{3}} + C$ $\displaystyle {{\cos x^3}\over{3}} + C$ $\displaystyle -{{\cos x^3}\over{3}} + C$
Evaluate $\displaystyle \int_{1}^{4}{x^{{{3}\over{2}}}\;dx}$ .
$\displaystyle\Big[ x^{{{5}\over{2}}} \Big]_1^4 = 31$ $\displaystyle\Big[ {{3\,\sqrt{x}}\over{2}} \Big]_1^4 = {{3}\over{2}}$ $\displaystyle\Big[ {{2\,x^{{{5}\over{2}}}}\over{5}} \Big]_1^4 = {{62}\over{5}}$ $\displaystyle\Big[ {{5\,x^{{{3}\over{2}}}}\over{2}} \Big]_1^4 = {{35}\over{2}}$ $\displaystyle\Big[ x^{{{3}\over{2}}} \Big]_1^4 = 7$
Find $\displaystyle \int {{{e^{x}}\over{\sqrt{e^{x}+1}}}}{\;dx}$ by substituting $\displaystyle u=e^{x}+1$
$\displaystyle \int {{{1}\over{\sqrt{u}}}}{\;du} = 2\,\sqrt{e^{x}+1} + C$ $\displaystyle \int {{{u-1}\over{\sqrt{u}}}}{\;du} = {{2\,\left(e^{x}+1\right)^{{{3}\over{2}}}}\over{3}}-2\,\sqrt{e^{x}+ 1} + C$ $\displaystyle \int {{{1}\over{u^{{{3}\over{2}}}}}}{\;du} = -{{2}\over{\sqrt{e^{x}+1}}} + C$ $\displaystyle -\int {{{1}\over{\sqrt{u}}}}{\;du} = -2\,\sqrt{e^{x}+1} + C$
Evaluate $\displaystyle \int_{-2}^{-1}{e^{x+2}\;dx}$ .
$\displaystyle\Big[ 2\,e^{x} \Big]_{ -2}^{ -1} = 2\,e^ {- 1 }-2\,e^ {- 2 }$ $\displaystyle\Big[ e^{x+2} \Big]_{ -2}^{ -1} = e-1$ $\displaystyle\Big[ e^{x} \Big]_{ -2}^{ -1} = e^ {- 1 }-e^ {- 2 }$ $\displaystyle\Big[ e^{x+3} \Big]_{ -2}^{ -1} = e^2-e$

Department of Mathematics
Last modified: 2026-08-28