Question

integrade x3.dx/(x+2)(x^2+3)​

Answer 1

Answer:

Use a substitution:

u=1+x2  

du=2xdx  

∫x3(1+x2)3dx=12∫x2(1+x2)3⋅2xdx=12∫(u−1)u3du=12∫(u−2−u−3)du=12(u−1−1−u−2−2)+C=14(−2u−1+u−2)+C=1−2u4u2+C=1−2(1+x2)4(1+x2)2+C=−1+2x24(1+x2)2+C  

Using trig substitution:

x=tanu  

dx=sec2u  

∫x3(1+x2)3dx=∫tan3u(1+tan2u)3⋅sec2udu=∫tan3u⋅sec2u(sec2u)3du=∫tan3usec4udu=∫sin3ucosudu=14sin4u+C=tan4u4sec4u+C=x44(1+x2)2+C  

NOTE:

Even though the results above seem different, they actually differ by just a constant:

(x44(1+x2)2)−(−1+2x24(1+x2)2)  

=x4+2x2+14(1+x2)2  

=(x2+1)24(1+x2)2  

=14  

Integration by parts:

u=x2du=2xdxdv=x(1+x2)3dxv=−14(1+x2)2  

∫x3(1+x2)3dx=−x24(1+x2)2+∫x2(1+x2)2dx=(14(1+x2)2−1−x24(1+x2)2)−14(1+x2)+C=14(1+x2)2−12(1+x2)+C  

This result is exactly the same as first one.

Step-by-step explanation: