Integrals involving ax+b

∫(ax+b)ndx=(ax+b)n+1a(n+1),(for n≠1)

∫1ax+bdx=1aln|ax+b|

∫x(ax+b)ndx=a(n+1)x−ba2(n+1)(n+2)(ax+b)n+1,(for n≠−1,n≠−2)

∫xax+bdx=x2−ba2ln|ax+b|

∫x(ax+b)2dx=ba2(ax+b)−1a2ln|ax+b|

∫x2ax+bdx=1a3((ax+b)22−2b(ax+b)+b2ln|ax+b|)

∫x2(ax+b)2dx=1a3(ax+b−2bln|ax+b|−b2ax+b)

∫x2(ax+b)3dx=1a3(ln|ax+b|+2bax+b−b22(ax+b)2)

∫x2(ax+b)ndx=1a3(−(ax+b)3−nn−3+2b(a+b)2−nn−2−b2(ax+b)1−nn−1)

∫1x(ax+b)dx=−1bln∣∣∣ax+bx∣∣∣

∫1x2(ax+b)2dx=−1bx+ab2ln∣∣∣ax+bx∣∣∣

∫1x2(ax+b)2dx=−a(1b2(ax+b)+1ab2x−2b3ln∣∣∣ax+bx∣∣∣)

Integrals involving ax2+bx+c

1x2+a2dx=1aarctanxa

1x2−a2dx=12aln∣∣∣x−ax+a∣∣∣

∫1ax2+bx+cdx=⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪24ac−b2−−−−−−−√arctan2ax+b4ac−b2−−−−−−−√for 4ac−b2>02b2−4ac−−−−−−−√ln∣∣∣∣2ax+b−b2−4ac−−−−−−−√2ax+b+b2−4ac−−−−−−−√∣∣∣∣for 4ac−b2<0−22ax+bfor 4ac−b2=0

∫xax2+bx+cdx=12aln∣∣ax2+bx+c∣∣−b2a∫dxax2+bx+c

∫1(ax2+bx+c)ndx=2ax+b(n−1)(4ac−b2)(ax+bx+c)n−1+2(2n−3)a(n−1)(4ac−b2)∫dx(ax2+bx+c)n−1

∫1x(ax2+bx+c)dx=12cln∣∣∣x2ax2+bx+c∣∣∣−b2c∫1ax2+bx+cdx