Calculus I - Computing Indefinite Integrals (Assignment Problems)

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Section 5.2 : Computing Indefinite Integrals

For problems 1 – 43 evaluate the given integral.

$\displaystyle \int{{7{x^5} - 5{x^4} + 6{x^2} - 14x + 3\,dx}}$
$\displaystyle \int{{{t^4} - 9{t^3} + 12{t^2} - 7t\,dt}}$
$\displaystyle \int{{4 - 18{w^{11}} - 9{w^9} + 8{w^7} + 2{w^5}\,dw}}$
$\displaystyle \int{{{x^9} - 6{x^4} - 21{x^2} - 1 + 9x\,dx}}$
$\displaystyle \int{{ - 7\,dz}}$
$\displaystyle \int{{4\,dw}}$
$\displaystyle \int{{10{z^{ - 6}} + 8{z^{ - 5}} - {z^{ - 2}} + 1\,dz}}$
$\displaystyle \int{{{y^{ - 16}} + 24{y^{ - 12}} - 14{y^{ - 8}} - 2{y^{ - 4}}\,dy}}$
$\displaystyle \int{{2{x^{ - 9}} + 12{x^{ - 5}} + 7{x^{ - 3}} - {x^{ - 2}}\,dx}}$
$\displaystyle \int{{5{z^{ - 4}} + 5{z^4} - \,9dz}}$
$\displaystyle \int{{6{t^3} + 8{t^{ - 6}} + {t^{ - 10}}\,dt}}$
$\displaystyle \int{{{x^{ - 3}} + 9{x^2} + 11{x^8} - 7{x^{ - 12}}\,dx}}$
$\displaystyle \int{{\sqrt[7]{{{w^2}}} + 3 - 9\,\,\sqrt[3]{{{w^7}}}\,\,dw}}$
$\displaystyle \int{{{w^5} + \sqrt {{w^5}} - \sqrt[5]{w}\,dw}}$
$\displaystyle \int{{6\,\,\sqrt[3]{{{v^2}}}}} - 7\,\,\sqrt[4]{v}\,dv$
$\displaystyle \int{{\frac{6}{{{y^3}}} - \frac{1}{{7{y^6}}} + \frac{1}{{{y^2}}}\,dy}}$
$\displaystyle \int{{8 + {u^5} - \frac{1}{{{u^5}}} + \frac{1}{{6{u^5}}}\,du}}$
$\displaystyle \int{{\frac{{12}}{{{x^5}}} + \frac{1}{{4{x^8}}} + \frac{6}{{7{x^2}}}\,dx}}$
$\displaystyle \int{{\sqrt[3]{{{t^5}}} - \frac{1}{{\sqrt {{t^9}} }} + {t^4}\,dt}}$
$\displaystyle \int{{\frac{2}{{{z^6}}} - \frac{1}{{5\,\,\sqrt[7]{{{z^8}}}}} + 9\,dz}}$
$\displaystyle \int{{{x^3} + \frac{1}{{{x^3}}} - \sqrt {{x^3}} \,dx}}$
$\displaystyle \int{{{x^6}\left( {1 - 4{x^2} + {x^3}} \right)\,dx}}$
$\displaystyle \int{{{{\left( {6 - 2u} \right)}^2}\,du}}$
$\displaystyle \int{{2 - \left( {3 + y} \right)\left( {4 - {y^3}} \right)\,dy}}$
$\displaystyle \int{{\sqrt w \left( {\sqrt[3]{w} - \sqrt[4]{w}} \right)\,dw}}$
$\displaystyle \int{{3v\left( {{v^2} - \frac{1}{{6{v^2}}} + \sqrt[3]{{{v^2}}}} \right)}}\,dv$
$\displaystyle \int{{\frac{{8{x^5} - 2{x^3} + 7}}{{{x^2}}}\,dx}}$
$\displaystyle \int{{\frac{{9 - z + 2{z^4} + 10{z^6}}}{{{z^4}}}\,dz}}$
$\displaystyle \int{{\frac{{2\sqrt t - 4t + \sqrt[3]{t}}}{{{t^2}}}\,dt}}$
$\displaystyle \int{{\frac{{\left( {1 - x} \right)\left( {2 + x} \right)}}{x}\,dx}}$
$\displaystyle \int{{6\sin \left( t \right) - 2\cos \left( t \right)\,dt}}$
$\displaystyle \int{{{{\sec }^2}\left( u \right) + 7\sec \left( u \right)\tan \left( u \right)\,du}}$
$\displaystyle \int{{{{\csc }^2}\left( y \right) - {{\sec }^2}\left( y \right)\,dy}}$
$\displaystyle \int{{8\cos \left( z \right) - 3\csc \left( z \right)\cot \left( z \right)\,dz}}$
$\displaystyle \int{{\tan \left( x \right)\left[ {\cot \left( x \right) - \cos \left( x \right)} \right]\,dx}}$
$\displaystyle \int{{\frac{{{{\cos }^3}\left( v \right) + \sin \left( v \right)}}{{{{\cos }^2}\left( v \right)}}\,dv}}$
$\displaystyle \int{{{w^2} + 2{{\bf{e}}^w}\,dw}}$
$\displaystyle \int{{{{\bf{e}}^t} + \frac{2}{t}\,dt}}$
$\displaystyle \int{{\frac{{14}}{x} - \frac{3}{{{x^2}}}\,dx}}$
$\displaystyle \int{{{{\bf{e}}^{ - u}}\left( {{{\bf{e}}^{2u}} + {{\bf{e}}^u}} \right)\,du}}$
$\displaystyle \int{{\frac{1}{{7z}} + \frac{1}{{{{\bf{e}}^{ - z}}}} + \frac{1}{{4{z^8}}}\,dz}}$
$\displaystyle \int{{1 + {w^2} - \frac{6}{{1 + {w^2}}}\,dw}}$
$\displaystyle \int{{\frac{5}{{1 + {t^2}}} + \frac{1}{{10\sqrt {1 - {t^2}} }}\,dt}}$
Determine $f\left( x \right)$ given that $f'\left( x \right) = 12{x^5} + 30{x^2}$ and $f\left( 4 \right) = - 23$ .
Determine $h\left( z \right)$ given that $h'\left( z \right) = 12{z^3} - 14{z^2} + 10$ and $h\left( { - 1} \right) = 8$ .
Determine $g\left( v \right)$ given that $\displaystyle g'\left( v \right) = \frac{1}{2}{v^{ - \,\frac{1}{2}}} - \frac{1}{4}{v^{ - \,\frac{3}{4}}}$ and $g\left( {16} \right) = 1$ .
Determine $P\left( t \right)$ given that $P'\left( t \right) = 6{{\bf{e}}^t} - 4 - 10t$ and $P\left( 0 \right) = - 6$ .
Determine $g\left( x \right)$ given that $g''\left( x \right) = 12{x^2} - 30x + 4$ , $g\left( { - 1} \right) = 7$ and $g\left( 2 \right) = 3$ .
Determine $f\left( u \right)$ given that $f''\left( u \right) = 60{u^4} - 60{u^2}$ , $f\left( { - 1} \right) = 14$ and $f'\left( 1 \right) = 6$ .
Determine $h\left( t \right)$ given that $h''\left( t \right) = 6t - 14 + 9{{\bf{e}}^t}$ , $h\left( 0 \right) = 4$ and $h\left( 3 \right) = 9{{\bf{e}}^3} + 8$ .