Exercice 9. Calculer les primitives suivantes par intégration par parties :
∫xexdx\int x e^{x}\,dx Posons u=x,;dv=exdxu=x,;dv=e^{x}dx. Alors du=dx,;v=exdu=dx,;v=e^{x}.
∫xexdx=xex−∫exdx=ex(x−1)+C.\int x e^{x}\,dx = x e^{x} - \int e^{x}\,dx = e^{x}(x-1)+C.
∫tsintdt\int t\sin t\,dt u=t,;dv=sint,dtu=t,;dv=\sin t,dt.
∫tsint,dt=−tcost+sint+C.\int t\sin t,dt = -t\cos t + \sin t + C.
∫lnxdx\int \ln x\,dx u=lnx,;dv=dxu=\ln x,;dv=dx.
∫lnxdx=xlnx−x+C.\int \ln x\,dx = x\ln x - x + C.
∫ln(2s+3)ds\int \ln(2s+3)\,ds u=ln(2s+3),;dv=dsu=\ln(2s+3),;dv=ds.
Après simplifications :
∫ln(2s+3)ds=(s+32)ln(2s+3)−s+C.\int \ln(2s+3)\,ds = \left(s+\frac{3}{2}\right)\ln(2s+3) - s + C.
∫xcosxdx\int x\cos x\,dx
∫xcosxdx=xsinx+cosx+C.\int x\cos x\,dx = x\sin x + \cos x + C.
∫ln(1+u2)du\int \ln(1+u^{2})\,du
∫ln(1+u2)du=uln(1+u2)−2u+2arctanu+C.\int \ln(1+u^{2})\,du = u\ln(1+u^{2}) - 2u + 2\arctan u + C.
∫arctanxdx\int \arctan x\,dx
∫arctanxdx=xarctanx−12ln(1+x2)+C.\int \arctan x\,dx = x\arctan x - \frac12\ln(1+x^{2}) + C.
∫ln2sds\int \ln^{2}s\,ds
∫ln2sds=s(ln2s−2lns+2)+C.\int \ln^{2}s\,ds = s\left(\ln^{2}s - 2\ln s + 2\right) + C.
∫lnxx2,dx\int \frac{\ln x}{x^{2}},dx
∫lnxx2dx=−lnx+1x+C.\int \frac{\ln x}{x^{2}}\,dx = -\frac{\ln x + 1}{x} + C.
∫ulnudu\int u\ln u\,du
∫ulnudu=u22lnu−u24+C.\int u\ln u\,du = \frac{u^{2}}{2}\ln u - \frac{u^{2}}{4} + C.
∫exsinxdx\int e^{x}\sin x\,dx
On trouve :
∫exsinxdx=ex2(sinx−cosx)+C.\int e^{x}\sin x\,dx = \frac{e^{x}}{2}(\sin x - \cos x) + C.