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

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

  2. Find the derivative $f'(x)$ for $f(x) = \displaystyle \int_{x}^{4}{t^2\,\sin t\;dt}$.

    $f'(x) = \displaystyle -x^2\,\cos x$ $f'(x) = \displaystyle -x^2\,\sin x$ $f'(x) = \displaystyle x^2\,\sin x$ $f'(x) = \displaystyle x^2\,\cos x$

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

    $\displaystyle -{{\cos x^2}\over{2}} + C$ $\displaystyle -{{\sin x^2}\over{2}} + C$ $\displaystyle {{\cos x^2}\over{2}} + C$ $\displaystyle {{\sin x^2}\over{2}} + C$

  4. Find $\displaystyle \int {\cos x\,\sin ^5x}{\;dx}$ by substituting $\displaystyle u=\sin x$

    $\displaystyle \int {u^5}{\;du} = {{\sin ^6x}\over{6}} + C$ $\displaystyle \int {u^5\,\sqrt{1-u^2}}{\;du} = -{{\sin ^4x\,\left(1-\sin ^2... ...{2}}}}\over{35}}-{{8\, \left(1-\sin ^2x\right)^{{{3}\over{2}}}}\over{105}} + C$ $\displaystyle -\int {u^5}{\;du} = -{{\sin ^6x}\over{6}} + C$ $\displaystyle \int {u^4}{\;du} = {{\sin ^5x}\over{5}} + C$

  5. Find $\displaystyle \int {\left(\sec x\right)^2\,\tan x}{\;dx}$ by using substitution.

    $\displaystyle -{{1}\over{\cos x}} + C$ $\displaystyle \ln \cos x + C$ $\displaystyle {{\tan ^2x}\over{2}} + C$ $\displaystyle {{1}\over{\cos x}} + C$

  6. Find $\displaystyle \int {{{e^{\sqrt{x}}}\over{\sqrt{x}}}}{\;dx}$ by using substitution.

    $\displaystyle -x + C$ $\displaystyle x + C$ $\displaystyle -2\,e^{\sqrt{x}} + C$ $\displaystyle 2\,e^{\sqrt{x}} + C$

  7. Evaluate $\displaystyle \int_{0}^{{{\pi}\over{4}}}{{{1-\cos ^2x}\over{\cos ^2x}}\;dx}$.

    $\displaystyle\Big[ -\tan x \Big]_{ 0}^{ {{\pi}\over{4}}} = -1$ $\displaystyle\Big[ x-\sin x \Big]_{ 0}^{ {{\pi}\over{4}}} = {{\sqrt{2}\,\pi-4}\over{2^{{{5}\over{2}}}}}$ $\displaystyle\Big[ {{1}\over{\cos ^2x}}-1 \Big]_{ 0}^{ {{\pi}\over{4}}} = 1$ $\displaystyle\Big[ \tan x-x \Big]_{ 0}^{ {{\pi}\over{4}}} = -{{\pi-4}\over{4}}$

  8. Evaluate $\displaystyle \int_{1}^{4}{{{1-x^2}\over{\sqrt{x}}}\;dx}$.

    $\displaystyle\Big[ 2\,\sqrt{x} \Big]_1^4 = 2$ $\displaystyle\Big[ x-{{x^3}\over{3}} \Big]_1^4 = -18$ $\displaystyle\Big[ 2\,\sqrt{x}-{{2\,x^{{{5}\over{2}}}}\over{5}} \Big]_1^4 = -{{52}\over{5}}$ $\displaystyle\Big[ {{\left(1-x^2\right)^2}\over{x}} \Big]_1^4 = {{225}\over{4}}$

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

    $\displaystyle \int {{{u+1}\over{\sqrt{u}}}}{\;du} = {{2\,\left(x-1\right)^{{{3}\over{2}}}}\over{3}}+2\,\sqrt{x-1} + C$ $\displaystyle \int {{{u-1}\over{\sqrt{u}}}}{\;du} = {{2\,\left(x+1\right)^{{{3}\over{2}}}}\over{3}}-2\,\sqrt{x+1} + C$ $\displaystyle \int {{{\left(u-1\right)^2}\over{\sqrt{u}}}}{\;du} = {{2\,\le... ...}}\over{5}}-{{4\,\left(x+1 \right)^{{{3}\over{2}}}}\over{3}}+2\,\sqrt{x+1} + C$ $\displaystyle \int {{{\left(u+1\right)^2}\over{\sqrt{u}}}}{\;du} = {{2\,\le... ...}}\over{5}}+{{4\,\left(x-1 \right)^{{{3}\over{2}}}}\over{3}}+2\,\sqrt{x-1} + C$

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

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


Department of Mathematics
Last modified: 2026-08-02