Find $\displaystyle \int {x^3\,\sqrt{x^4+3}}{\;dx}$ by substituting $\displaystyle u=x^4+3$
$\displaystyle \int {\sqrt{u}}{\;du} = {{2\,\left(x^4+3\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{\int {\left(u-3\right)\,\sqrt{u}}{\;du}}\over{4}} = {{\left... ...{{{5}\over{2}}}}\over{10}}-{{\left(x^4+3 \right)^{{{3}\over{2}}}}\over{2}} + C$ $\displaystyle {{\int {\sqrt{u}}{\;du}}\over{4}} = {{\left(x^4+3\right)^{{{3}\over{2}}}}\over{6}} + C$ $\displaystyle \int {{{1}\over{\sqrt{u}}}}{\;du} = 2\,\sqrt{x^4+3} + C$
Evaluate $\displaystyle \int_{1}^{4}{{{\sqrt{x}+1}\over{\sqrt{x}}}\;dx}$ .
$\displaystyle\Big[ x+2\,\sqrt{x} \Big]_1^4 = 5$ $\displaystyle\Big[ {{\left(\sqrt{x}+1\right)^2}\over{x}} \Big]_1^4 = -{{7}\over{4}}$ $\displaystyle\Big[ {{2\,x^{{{3}\over{2}}}}\over{3}}+x \Big]_1^4 = {{23}\over{3}}$ $\displaystyle\Big[ 2\,\sqrt{x} \Big]_1^4 = 2$
Evaluate $\displaystyle \int_{1}^{4}{x^{{{3}\over{2}}}\;dx}$ .
$\displaystyle\Big[ {{3\,\sqrt{x}}\over{2}} \Big]_1^4 = {{3}\over{2}}$ $\displaystyle\Big[ x^{{{5}\over{2}}} \Big]_1^4 = 31$ $\displaystyle\Big[ x^{{{3}\over{2}}} \Big]_1^4 = 7$ $\displaystyle\Big[ {{5\,x^{{{3}\over{2}}}}\over{2}} \Big]_1^4 = {{35}\over{2}}$ $\displaystyle\Big[ {{2\,x^{{{5}\over{2}}}}\over{5}} \Big]_1^4 = {{62}\over{5}}$
Find $\displaystyle \int {{{x^2+1}\over{\sqrt{x}}}}{\;dx}$ .
$\displaystyle 2\,\sqrt{x} + C$ $\displaystyle {{\left(x^2+1\right)^2}\over{x}} + C$ $\displaystyle {{x^3}\over{3}}+x + C$ $\displaystyle {{2\,x^{{{5}\over{2}}}+10\,\sqrt{x}}\over{5}} + C$
Find $\displaystyle \int {{{\cos \sqrt{x}}\over{\sqrt{x}}}}{\;dx}$ by using substitution.
$\displaystyle -2\,\sin \sqrt{x} + C$ $\displaystyle 2\,\sin \sqrt{x} + C$ $\displaystyle -2\,\cos \sqrt{x} + C$ $\displaystyle 2\,\cos \sqrt{x} + C$
Find $\displaystyle \int {\left(\sec x\right)^2\,\tan x}{\;dx}$ by using substitution.
$\displaystyle \ln \cos x + C$ $\displaystyle {{\tan ^2x}\over{2}} + C$ $\displaystyle {{1}\over{\cos x}} + C$ $\displaystyle -{{1}\over{\cos x}} + C$
Find $\displaystyle -\int {\cos x\,\sin ^5x}{\;dx}$ by substituting $\displaystyle u=\sin x$
$\displaystyle -\int {u^4}{\;du} = -{{\sin ^5x}\over{5}} + C$ $\displaystyle -\int {u^5\,\sqrt{1-u^2}}{\;du} = {{\sin ^4x\,\left(1-\sin ^2... ...{2}}}}\over{35}}+{{8\, \left(1-\sin ^2x\right)^{{{3}\over{2}}}}\over{105}} + C$ $\displaystyle \int {u^5}{\;du} = {{\sin ^6x}\over{6}} + C$ $\displaystyle -\int {u^5}{\;du} = -{{\sin ^6x}\over{6}} + C$
Evaluate $\displaystyle \int_{0}^{1}{\left(4-x\right)^2\;dx}$ .
$\displaystyle\Big[ 4\,x-{{x^2}\over{2}} \Big]_0^1 = {{7}\over{2}}$ $\displaystyle\Big[ {{x^3}\over{3}}-4\,x^2+16\,x \Big]_0^1 = {{37}\over{3}}$ $\displaystyle\Big[ {{x^2}\over{2}} \Big]_0^1 = {{1}\over{2}}$ $\displaystyle\Big[ x^2-8\,x+16 \Big]_0^1 = -7$
Find $\displaystyle \int {1-{{1}\over{x^2+1}}}{\;dx}$ .
$\displaystyle x-\arctan x + C$ $\displaystyle x+{{1}\over{x}} + C$ $\displaystyle {{\ln \left(x+1\right)}\over{2}}+x-{{\ln \left(x-1\right)}\over{2 }} + C$ $\displaystyle {{2\,x}\over{\left(x^2+1\right)^2}} + C$
Find the derivative for $f(x) = \displaystyle \int_{0}^{x}{t\,\sin ^2t\;dt}$ .
$f'(x) = \displaystyle -x\,\cos ^2x$ $f'(x) = \displaystyle -x\,\sin ^2x$ $f'(x) = \displaystyle x\,\sin ^2x$ $f'(x) = \displaystyle x\,\cos ^2x$

Department of Mathematics
Last modified: 2025-08-29