# Calculus/Integration techniques/Integration by Parts/Solutions

1. ${\displaystyle \int -4\ln(x)dx}$
Let
${\displaystyle {\begin{array}{ll}u=\ln(x)&\qquad du={\frac {dx}{x}}\\v=-4x&\qquad dv=-4dx\end{array}}}$

Then

{\displaystyle {\begin{aligned}\int -4\ln(x)dx&=-4x\ln(x)-\int -4dx\\&=\mathbf {-4x\ln(x)+4x+C} \end{aligned}}}
Let
${\displaystyle {\begin{array}{ll}u=\ln(x)&\qquad du={\frac {dx}{x}}\\v=-4x&\qquad dv=-4dx\end{array}}}$

Then

{\displaystyle {\begin{aligned}\int -4\ln(x)dx&=-4x\ln(x)-\int -4dx\\&=\mathbf {-4x\ln(x)+4x+C} \end{aligned}}}
2. ${\displaystyle \int (-7x+38)\cos(x)dx}$
Let
${\displaystyle {\begin{array}{ll}u=-7x+38&\qquad du=-7dx\\v=\sin(x)&\qquad dv=\cos(x)dx\end{array}}}$

Then

{\displaystyle {\begin{aligned}\int (-7x+38)\cos(x)dx&=(-7x+38)\sin(x)+7\int \sin(x)dx\\&=\mathbf {(-7x+38)\sin(x)-7\cos(x)+C} \end{aligned}}}
Let
${\displaystyle {\begin{array}{ll}u=-7x+38&\qquad du=-7dx\\v=\sin(x)&\qquad dv=\cos(x)dx\end{array}}}$

Then

{\displaystyle {\begin{aligned}\int (-7x+38)\cos(x)dx&=(-7x+38)\sin(x)+7\int \sin(x)dx\\&=\mathbf {(-7x+38)\sin(x)-7\cos(x)+C} \end{aligned}}}
3. ${\displaystyle \int \limits _{0}^{\frac {\pi }{2}}(-6x+45)\cos(x)dx}$
Let
${\displaystyle {\begin{array}{ll}u=-6x+45&\qquad du=-6dx\\v=\sin(x)&\qquad dv=\cos(x)dx\end{array}}}$

Then

{\displaystyle {\begin{aligned}\int \limits _{0}^{\frac {\pi }{2}}(-6x+45)\cos(x)dx&=((-6x+45)\sin(x)){\bigg |}_{0}^{\frac {\pi }{2}}+6\int \limits _{0}^{\frac {\pi }{2}}\sin(x)dx\\&=(-3\pi +45)-6\cos(x){\bigg |}_{0}^{\frac {\pi }{2}}\\&=(-3\pi +45)+6\\&=\mathbf {-3\pi +51} \end{aligned}}}
Let
${\displaystyle {\begin{array}{ll}u=-6x+45&\qquad du=-6dx\\v=\sin(x)&\qquad dv=\cos(x)dx\end{array}}}$

Then

{\displaystyle {\begin{aligned}\int \limits _{0}^{\frac {\pi }{2}}(-6x+45)\cos(x)dx&=((-6x+45)\sin(x)){\bigg |}_{0}^{\frac {\pi }{2}}+6\int \limits _{0}^{\frac {\pi }{2}}\sin(x)dx\\&=(-3\pi +45)-6\cos(x){\bigg |}_{0}^{\frac {\pi }{2}}\\&=(-3\pi +45)+6\\&=\mathbf {-3\pi +51} \end{aligned}}}
4. ${\displaystyle \int (5x+1)(x-6)^{4}dx}$
Let
${\displaystyle {\begin{array}{ll}u=5x+1&\qquad du=5dx\\v={\frac {(x-6)^{5}}{5}}&\qquad dv=(x-6)^{4}dx\end{array}}}$

Then

{\displaystyle {\begin{aligned}\int (5x+1)(x-6)^{4}dx&={\frac {(5x+1)(x-6)^{5}}{5}}-\int (x-6)^{5}dx\\&=\mathbf {{\frac {(5x+1)(x-6)^{5}}{5}}-{\frac {(x-6)^{6}}{6}}+C} \end{aligned}}}
Let
${\displaystyle {\begin{array}{ll}u=5x+1&\qquad du=5dx\\v={\frac {(x-6)^{5}}{5}}&\qquad dv=(x-6)^{4}dx\end{array}}}$

Then

{\displaystyle {\begin{aligned}\int (5x+1)(x-6)^{4}dx&={\frac {(5x+1)(x-6)^{5}}{5}}-\int (x-6)^{5}dx\\&=\mathbf {{\frac {(5x+1)(x-6)^{5}}{5}}-{\frac {(x-6)^{6}}{6}}+C} \end{aligned}}}
5. ${\displaystyle \int \limits _{-1}^{1}(2x+8)^{3}(-x+2)dx}$
Let
${\displaystyle {\begin{array}{ll}u=-x+2&\qquad du=-dx\\v={\frac {(2x+8)^{4}}{8}}&\qquad dv=(2x+8)^{3}dx\end{array}}}$

Then

{\displaystyle {\begin{aligned}\int \limits _{-1}^{1}(2x+8)^{3}(-x+2)dx&={\frac {(-x+2)(2x+8)^{4}}{8}}{\Bigg |}_{-1}^{1}+\int \limits _{-1}^{1}{\frac {(2x+8)^{4}}{8}}dx\\&={\frac {10^{4}-(3\cdot 6^{4})}{8}}+{\frac {1}{8}}{\frac {(2x+8)^{5}}{10}}{\Bigg |}_{-1}^{1}\\&={\frac {10^{4}-3\cdot 6^{4}}{8}}+{\frac {1}{80}}(10^{5}-6^{5})\\&={\frac {10^{5}-30\cdot 6^{4}+10^{5}-6^{5}}{80}}\\&={\frac {2\cdot 10^{5}-6^{4}(30+6)}{80}}\\&={\frac {2\cdot 10^{5}-6^{6}}{80}}\\&=\mathbf {\frac {9584}{5}} \end{aligned}}}
Let
${\displaystyle {\begin{array}{ll}u=-x+2&\qquad du=-dx\\v={\frac {(2x+8)^{4}}{8}}&\qquad dv=(2x+8)^{3}dx\end{array}}}$

Then

{\displaystyle {\begin{aligned}\int \limits _{-1}^{1}(2x+8)^{3}(-x+2)dx&={\frac {(-x+2)(2x+8)^{4}}{8}}{\Bigg |}_{-1}^{1}+\int \limits _{-1}^{1}{\frac {(2x+8)^{4}}{8}}dx\\&={\frac {10^{4}-(3\cdot 6^{4})}{8}}+{\frac {1}{8}}{\frac {(2x+8)^{5}}{10}}{\Bigg |}_{-1}^{1}\\&={\frac {10^{4}-3\cdot 6^{4}}{8}}+{\frac {1}{80}}(10^{5}-6^{5})\\&={\frac {10^{5}-30\cdot 6^{4}+10^{5}-6^{5}}{80}}\\&={\frac {2\cdot 10^{5}-6^{4}(30+6)}{80}}\\&={\frac {2\cdot 10^{5}-6^{6}}{80}}\\&=\mathbf {\frac {9584}{5}} \end{aligned}}}