En utilisant que ${\displaystyle \int {\frac {\ln ^{k}x}{x^{2}}}\,\mathrm {d} x=\left[{\frac {-\ln ^{k}x}{x}}\right]+k\int {\frac {\ln ^{k-1}x}{x^{2}}}\,\mathrm {d} x}$ pour ${\displaystyle k=3,2,1,0}$, on trouve une primitive de ${\displaystyle f}$ sur ${\displaystyle \mathbb {R} _{+}^{*}}$ :

${\displaystyle F(x)={\frac {-1}{x}}\left(\ln ^{3}x+3\ln ^{2}x+6\ln x+6\right)}$.

Plus généralement, une primitive sur ${\displaystyle \mathbb {R} _{+}^{*}}$ de ${\displaystyle f(x)={\frac {\ln ^{n}x}{x^{2}}}}$ est ${\displaystyle F(x)={\frac {-1}{x}}\sum _{k=0}^{n}{\frac {n!}{k!}}\ln ^{k}x}$.

${\displaystyle {\begin{aligned}\int f&=\left[\left(x^{3}+x\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]+{\frac {1}{3}}\int \left(3x^{2}+1\right)\mathrm {e} ^{-3x}\,\mathrm {d} x\\&=\left[\left(x^{3}+x+{\frac {3x^{2}+1}{3}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]+{\frac {2}{3}}\int x\mathrm {e} ^{-3x}\,\mathrm {d} x\\&=\left[\left(x^{3}+x+{\frac {3x^{2}+1}{3}}+{\frac {2x}{3}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]+{\frac {2}{9}}\int \mathrm {e} ^{-3x}\,\mathrm {d} x\\&=\left[\left(x^{3}+x+{\frac {3x^{2}+1}{3}}+{\frac {2x}{3}}+{\frac {2}{9}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]\\&=\left[\left(x^{3}+x^{2}+{\frac {5x}{3}}+{\frac {5}{9}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]\end{aligned}}}$

donc une primitive de ${\displaystyle f}$ sur ${\displaystyle \mathbb {R} }$ est ${\displaystyle F:x\mapsto \left(x^{3}+x^{2}+{\frac {5x}{3}}+{\frac {5}{9}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}}$.

${\displaystyle {\begin{aligned}\int f&=\left[\left(x^{2}-x+6\right){\frac {-\mathrm {e} ^{-5x}}{5}}\right]+{\frac {1}{5}}\int \left(2x-1\right)\mathrm {e} ^{-5x}\,\mathrm {d} x\\&=\left[\left(x^{2}-x+6+{\frac {2x-1}{5}}\right){\frac {-\mathrm {e} ^{-5x}}{5}}\right]+{\frac {2}{25}}\int \mathrm {e} ^{-5x}\,\mathrm {d} x\\&=\left[\left(x^{2}-x+6+{\frac {2x-1}{5}}+{\frac {2}{25}}\right){\frac {-\mathrm {e} ^{-5x}}{5}}\right]\\&=\left[\left(x^{2}-{\frac {3x}{5}}+{\frac {147}{25}}\right){\frac {-\mathrm {e} ^{-5x}}{5}}\right]\end{aligned}}}$

donc une primitive de ${\displaystyle f}$ sur ${\displaystyle \mathbb {R} }$ est ${\displaystyle F:x\mapsto \left(x^{2}-{\frac {3x}{5}}+{\frac {147}{25}}\right){\frac {-\mathrm {e} ^{-5x}}{5}}}$.

Sur chaque intervalle ${\displaystyle \left]k\pi -{\frac {\pi }{2}},k\pi +{\frac {\pi }{2}}\right[}$ (avec ${\displaystyle k\in \mathbb {Z} }$), une primitive de ${\displaystyle f=\tan \times \tan '}$ est ${\displaystyle F={\frac {\tan ^{2}}{2}}}$.

${\displaystyle f(x)=-{\frac {u'\mathrm {e} ^{u}}{3}}}$ avec ${\displaystyle u(x)={\frac {1}{x^{3}}}}$ donc sur chacun des deux intervalles ${\displaystyle \left]-\infty ,0\right[}$ et ${\displaystyle \left]0,+\infty \right[}$, une primitive de ${\displaystyle f}$ est ${\displaystyle F=-{\frac {\mathrm {e} ^{u}}{3}}}$.

${\displaystyle {\begin{aligned}\int f&=\left[\cos 2x\mathrm {e} ^{x}\right]+2\int \sin 2x\mathrm {e} ^{x}\,\mathrm {d} x\\&=\left[\left(\cos 2x+2\sin 2x\right)\mathrm {e} ^{x}\right]-4\int f\end{aligned}}}$

donc une primitive de ${\displaystyle f}$ sur ${\displaystyle \mathbb {R} }$ est ${\displaystyle F:x\mapsto {\frac {\left(\cos 2x+2\sin 2x\right)\mathrm {e} ^{x}}{5}}}$.

${\displaystyle {\begin{aligned}\int f&=\left[{\frac {\mathrm {e} ^{2x}}{2}}\sin 2x\right]-\int \mathrm {e} ^{2x}\cos 2x\,\mathrm {d} x\\&=\left[{\frac {\mathrm {e} ^{2x}}{2}}\left(\sin 2x-\cos 2x\right)\right]-\int f\end{aligned}}}$

donc une primitive de ${\displaystyle f}$ sur ${\displaystyle \mathbb {R} }$ est ${\displaystyle F:x\mapsto {\frac {\mathrm {e} ^{2x}\left(\sin 2x-\cos 2x\right)}{4}}}$.

${\displaystyle {\begin{aligned}\int f&=\left[{\frac {\mathrm {e} ^{3x}}{3}}\left(\sin 2x-\cos 2x\right)\right]-{\frac {2}{3}}\int \mathrm {e} ^{3x}\left(\cos 2x+\sin 2x\right)\,\mathrm {d} x\\&=\left[{\frac {\mathrm {e} ^{3x}}{3}}\left(\sin 2x-\cos 2x-{\frac {2}{3}}\left(\cos 2x+\sin 2x\right)\right)\right]+{\frac {4}{9}}\int \mathrm {e} ^{3x}\left(-\sin 2x+\cos 2x\right)\,\mathrm {d} x\\&=\left[{\frac {\mathrm {e} ^{3x}\left(\sin 2x-5\cos 2x\right)}{9}}\right]-{\frac {4}{9}}\int f\end{aligned}}}$

donc une primitive de ${\displaystyle f}$ sur ${\displaystyle \mathbb {R} }$ est ${\displaystyle F:x\mapsto {\frac {\mathrm {e} ^{3x}\left(\sin 2x-5\cos 2x\right)}{13}}}$.

Sur ${\displaystyle \left]-1,+\infty \right[}$,

${\displaystyle \int f=\left[{\frac {x^{3}}{3}}\ln(x+1)\right]-{\frac {1}{3}}\int {\frac {x^{3}}{x+1}}\,\mathrm {d} x}$ et

${\displaystyle {\frac {x^{3}}{x+1}}={\frac {x^{3}+1}{x+1}}-{\frac {1}{x+1}}=x^{2}-x+1-{\frac {1}{x+1}}}$

donc une primitive de ${\displaystyle f}$ sur ${\displaystyle \left]-1,+\infty \right[}$ est

${\displaystyle F:x\mapsto {\frac {1}{3}}\left(\left(x^{3}+1\right)\ln(x+1)-{\frac {x^{3}}{3}}+{\frac {x^{2}}{2}}-x\right)}$.

${\displaystyle \int f=\left[x\cos \left(\ln x\right)\right]+\int g}$ et ${\displaystyle \int g=\left[x\sin \left(\ln x\right)\right]-\int f}$

donc (sur ${\displaystyle \left]0,+\infty \right[}$) une primitive de ${\displaystyle f}$ est

${\displaystyle F:x\mapsto {\frac {x}{2}}\left(\cos \left(\ln x\right)+\sin \left(\ln x\right)\right)}$

et une primitive de ${\displaystyle g}$ est

${\displaystyle G:x\mapsto {\frac {x}{2}}\left(\sin \left(\ln x\right)-\cos \left(\ln x\right)\right)}$.

${\displaystyle {\begin{aligned}\int f&=\left[x\ln ^{2}x\right]-2\int \ln x\,\mathrm {d} x\\&=\left[x\ln ^{2}x-2x\ln x\right]+2\int \mathrm {d} x\end{aligned}}}$

donc une primitive de ${\displaystyle f}$ sur ${\displaystyle \left]0,+\infty \right[}$ est

${\displaystyle F:x\mapsto x\left(\ln ^{2}x-2\ln x+2\right)}$.

${\displaystyle f=\mathrm {e} ^{u}(u-1)u'}$ avec ${\displaystyle u(x)=x^{2}+1}$ donc une primitive de ${\displaystyle f}$ sur ${\displaystyle \mathbb {R} }$ est ${\displaystyle F=(u-2)\operatorname {e} ^{u}}$, c'est-à-dire ${\displaystyle F(x)=(x^{2}-1)\operatorname {e} ^{x^{2}+1}}$.

${\displaystyle \forall x>0\quad -\int _{x}^{+\infty }f(t)\;\mathrm {d} t=\left[{\frac {\ln(1+t^{2})}{t}}\right]_{x}^{+\infty }-\int _{x}^{+\infty }{\frac {2}{1+t^{2}}}\;\mathrm {d} t=\left[{\frac {\ln(1+t^{2})}{t}}-2\arctan t\right]_{x}^{+\infty }=-{\frac {\ln(1+x^{2})}{x}}-\pi +2\arctan x}$

donc (par parité de ${\displaystyle f}$) une primitive de ${\displaystyle f}$ sur ${\displaystyle \mathbb {R} ^{*}}$ est

${\displaystyle F(x)=-{\frac {\ln(1+x^{2})}{x}}+2\arctan x}$.

On en déduit :

${\displaystyle \forall y>0\quad \int _{0}^{y}g(s)\;\mathrm {d} s=\int _{1/y}^{+\infty }f(t)\;\mathrm {d} t=y\ln(1+1/y^{2})+\pi -2\arctan {\frac {1}{y}}=y\ln(1+1/y^{2})+2\arctan y}$

donc (par parité de ${\displaystyle g}$) une primitive de ${\displaystyle g}$ sur ${\displaystyle \mathbb {R} ^{*}}$ est

${\displaystyle G(y)=y\ln(1+1/y^{2})+2\arctan y}$.

1. ${\displaystyle {\frac {x^{4}}{2}}+x^{3}-x+c}$
2. ${\displaystyle {\frac {x^{{\frac {4}{5}}+1}}{{\frac {4}{5}}+1}}+c={\frac {5x^{\frac {9}{5}}}{9}}+c}$
3. Changement de variable ${\displaystyle x=y^{2}}$, ${\displaystyle \int f_{3}(x)\;\mathrm {d} x=\int (1-y^{2})y\;2y\mathrm {d} y={\frac {2y^{3}}{3}}-{\frac {2y^{5}}{5}}+c={\frac {2x^{\frac {3}{2}}}{3}}-{\frac {2x^{\frac {5}{2}}}{5}}+c=2{\sqrt {x}}\left({\frac {x}{3}}-{\frac {x^{2}}{5}}\right)+c}$
4. ${\displaystyle {\frac {(3x+2)^{4}}{12}}+c}$
5. ${\displaystyle {\frac {\sin(4x+1)}{4}}+c}$
6. Changement de variable ${\displaystyle y={\sqrt {1-x^{2}}}}$, ${\displaystyle \int f_{6}(x)\;\mathrm {d} x=-\int y^{2}\mathrm {d} y=-{\frac {y^{3}}{3}}+c=-{\frac {(1-x^{2})^{\frac {3}{2}}}{3}}+c}$
7. ${\displaystyle {\frac {(5\sin x+2)^{4}}{20}}+c}$
8. ${\displaystyle {\frac {\ln |3x+5|}{3}}+c_{\pm }}$ : ${\displaystyle c_{-}}$ sur ${\displaystyle \left]-\infty ,-{\frac {5}{3}}\right[}$, ${\displaystyle c_{+}}$ sur ${\displaystyle \left]-{\frac {5}{3}},+\infty \right[}$
9. Changement de variable ${\displaystyle y=\mathrm {e} ^{x}}$, ${\displaystyle \int f_{9}(x)\;\mathrm {d} x=\int {\frac {\mathrm {d} y}{y(1+y)}}=\int \left({\frac {1}{y}}-{\frac {1}{1+y}}\right)\;\mathrm {d} y=\ln |y|-\ln |1+y|+c=x-\ln(1+\mathrm {e} ^{x})+c}$
10. ${\displaystyle {\frac {(\mathrm {e} ^{2x}+2)^{5}}{10}}+c}$
11. ${\displaystyle \int \left(1-{\frac {1}{x+2}}\right)\;\mathrm {d} x=x-\ln |x+2|+c_{\pm }}$
12. ${\displaystyle {\frac {\arctan {\frac {x}{a}}}{a}}+c}$
13. ${\displaystyle {\frac {\ln(x^{2}+4)}{2}}+c}$
14. ${\displaystyle f_{14}(x)=1+{\frac {1}{x-2}}-{\frac {1}{x+2}}}$, ${\displaystyle F_{14}(x)=x+\ln |x-2|-\ln |x+2|+c_{j}}$
15. ${\displaystyle f_{15}(x)={\frac {1}{4}}\left({\frac {1}{x+1}}-{\frac {1}{x+5}}\right)}$, ${\displaystyle F_{15}(x)={\frac {1}{4}}\left(\ln |x+1|-\ln |x+5|\right)+c_{j}}$
16. ${\displaystyle f_{16}(x)={\frac {1-\cos(2x)}{2}}}$, ${\displaystyle F_{16}(x)={\frac {x}{2}}-{\frac {\sin(2x)}{4}}+c}$
17. ${\displaystyle f_{17}(x)={\frac {1}{8}}\cos(8x)+{\frac {1}{2}}\cos(4x)+{\frac {3}{8}}}$, ${\displaystyle F_{17}(x)={\frac {1}{64}}\sin(4x)+{\frac {1}{8}}\sin(2x)+{\frac {3x}{8}}+c}$
18. ${\displaystyle f_{18}(x)={\frac {1}{2}}\cos {\big (}(a+b)x{\big )}+{\frac {1}{2}}\cos {\big (}(a-b)x{\big )}}$, ${\displaystyle F_{18}(x)={\frac {1}{2(a+b)}}\sin {\big (}(a+b)x{\big )}+{\frac {1}{2(a-b)}}\sin {\big (}(a-b)x{\big )}+c}$
19. ${\displaystyle \left[x\arctan x\right]-\int {\frac {x}{1+x^{2}}}\;\mathrm {d} x=x\arctan x-{\frac {\ln(1+x^{2})}{2}}+c}$
20. ${\displaystyle \left[x\ln(x^{2}+2)\right]-2\int {\frac {x^{2}}{x^{2}+2}}\;\mathrm {d} x=\left[x\ln(x^{2}+2)\right]-2\int \left(1-frac2{x^{2}+2}\right)\;\mathrm {d} x=x\ln(x^{2}+2)-2x+2{\sqrt {2}}\arctan {\frac {x}{\sqrt {2}}}+c}$
21. ${\displaystyle \int \cos {\sqrt {x}}\;\mathrm {d} x=\int \cos s\;2s\mathrm {d} s}$. Une primitive de ${\displaystyle s\cos s}$ (cf. Intégration_en_mathématiques/Exercices/Primitives_3#Exercice_6-2) est ${\displaystyle s\sin s+\cos s}$. Donc ${\displaystyle F_{21}(x)=2({\sqrt {x}}\sin {\sqrt {x}}+\cos {\sqrt {x}})+c}$.
22. ${\displaystyle \left[{\frac {-\ln x}{1+x}}\right]+\int {\frac {1}{x(1+x)}}\;\mathrm {d} x=\left[{\frac {-\ln x}{1+x}}\right]+\int \left({\frac {1}{x}}-{\frac {1}{1+x}}\right)\;\mathrm {d} x}$ donc ${\displaystyle F_{22}(x)={\frac {-\ln x}{1+x}}+\ln {\frac {x}{1+x}}+c={\frac {x\ln x}{1+x}}-\ln(1+x)+c}$.