integration of x cube+x+1/xsquare-1
Answers
Answer:
Step-by-step explanation:
You can do a u-substitution.
Notice how this can be written as:
∫x2⋅(11−x3)dx
ddx[−x3]=−3x2dx, so ddx[(−13)⋅(−x3)]=x2dx.
Let:
u=1−x3
du=−3x2dx
⇒−13∫(11−x3)⋅(−3x2)dx
=−13∫1udu
=−13ln|u|+C
=−13ln∣∣1−x3∣∣+C
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