The following is a demo of aligning integral lists.
Indefinite integrals are antiderivative functions. A constant (the constant of integration ) may be added to the right hand side of any of these formulas, but has been suppressed here in the interest of brevity.
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{\displaystyle {\begin{aligned}\int xe^{cx}\;\mathrm {d} x&=e^{cx}\left({\frac {cx-1}{c^{2}}}\right)\\\int x^{2}e^{cx}\;\mathrm {d} x&=e^{cx}\left({\frac {x^{2}}{c}}-{\frac {2x}{c^{2}}}+{\frac {2}{c^{3}}}\right)\\\int x^{n}e^{cx}\;\mathrm {d} x&={\frac {1}{c}}x^{n}e^{cx}-{\frac {n}{c}}\int x^{n-1}e^{cx}\mathrm {d} x=\left({\frac {\partial }{\partial c}}\right)^{n}{\frac {e^{cx}}{c}}=e^{cx}\sum _{i=0}^{n}(-1)^{i}\,{\frac {n!}{(n-i)!\,c^{i+1}}}\,x^{n-i}=e^{cx}\sum _{i=0}^{n}(-1)^{n-i}\,{\frac {n!}{i!\,c^{n-i+1}}}\,x^{i}\\\int {\frac {e^{cx}}{x}}\;\mathrm {d} x&=\ln |x|+\sum _{n=1}^{\infty }{\frac {(cx)^{n}}{n\cdot n!}}\\\int {\frac {e^{cx}}{x^{n}}}\;\mathrm {d} x&={\frac {1}{n-1}}\left(-{\frac {e^{cx}}{x^{n-1}}}+c\int {\frac {e^{cx}}{x^{n-1}}}\,\mathrm {d} x\right)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\\\end{aligned}}}
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{\displaystyle \int xe^{cx}\;\mathrm {d} x=e^{cx}\left({\frac {cx-1}{c^{2}}}\right)}
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{\displaystyle \int x^{2}e^{cx}\;\mathrm {d} x=e^{cx}\left({\frac {x^{2}}{c}}-{\frac {2x}{c^{2}}}+{\frac {2}{c^{3}}}\right)}
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{\displaystyle \int x^{n}e^{cx}\;\mathrm {d} x={\frac {1}{c}}x^{n}e^{cx}-{\frac {n}{c}}\int x^{n-1}e^{cx}\mathrm {d} x=\left({\frac {\partial }{\partial c}}\right)^{n}{\frac {e^{cx}}{c}}=e^{cx}\sum _{i=0}^{n}(-1)^{i}\,{\frac {n!}{(n-i)!\,c^{i+1}}}\,x^{n-i}=e^{cx}\sum _{i=0}^{n}(-1)^{n-i}\,{\frac {n!}{i!\,c^{n-i+1}}}\,x^{i}}
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{\displaystyle \int {\frac {e^{cx}}{x}}\;\mathrm {d} x=\ln |x|+\sum _{n=1}^{\infty }{\frac {(cx)^{n}}{n\cdot n!}}}
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{\displaystyle \int {\frac {e^{cx}}{x^{n}}}\;\mathrm {d} x={\frac {1}{n-1}}\left(-{\frac {e^{cx}}{x^{n-1}}}+c\int {\frac {e^{cx}}{x^{n-1}}}\,\mathrm {d} x\right)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
