August 13, 2026 · Reference · LibreTimesTable of Basic IntegralsSergeyMathematical AnalysisBasic indefinite integrals of the elementary functions. ∫xndx=n+1xn+1+C,n=−1 ∫xdx=ln∣x∣+C ∫axdx=lnaax+C,a>0,a=1 ∫exdx=ex+C ∫cosxdx=sinx+C ∫sinxdx=−cosx+C ∫cos2xdx=tanx+C ∫sin2xdx=−cotx+C ∫a2−x2dx=arcsinax+C,a>0 ∫a2+x2dx=a1⋅arctanax+C,a=0 ∫x2−a2dx=2a1⋅lnx+ax−a+C,a=0 ∫x2±a2dx=lnx+x2±a2+C,a=0(*) ∫sinxdx=lntan2x+C(**) ∫cosxdx=lntan(2x+4π)+C ∫coshxdx=sinhx+C ∫sinhxdx=coshx+C ∫cosh2xdx=tanhx+C ∫sinh2xdx=−cothx+C ∫axdx=lna1ax+C ∫lnxdx=xlnx−x+C ∫tanxdx=ln∣secx∣+C ∫sec2xdx=tanx+C ∫secxtanxdx=secx+C ∫secxdx=ln∣secx+tanx∣+C ∫a2+x2adx=arctanax+C ∫a2−x21dx=arcsinax+C ∫xndx=n+11xn+1+C,n=−1 ∫x1dx=ln∣x∣+C ∫udv=uv−∫vdu ∫ax+b1dx=a1ln∣ax+b∣+C ∫xndx=n+11xn+1+C,n=−1 ∫xx2−a2adx=arcsecax+C ∫a2−x2adx=21lnx−ax+a+C ∫x2+a21dx=sinh−1ax=ln(x+x2+a2)+C ∫x2−a21dx=cosh−1ax=ln(x+x2−a2)+C
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