What is the antiderivative of x31+x2?

1 Answer
Jul 7, 2016

12{x2+1ln(x2+1)}+C.

OR

12{x2ln(x2+1)}+C1, where C1=C+12.

Explanation:

Let I=x31+x2dx

We take substn. x2+1=t, so that, 2xdx=dt.

Also, x2+1=tx2=t1

Now, I=x31+x2dx=12x22x1+x2dx=12t1tdt
=12{tt1t}dt=12{11t}dt=12{tlnt}=12{x2+1ln(x2+1)}+C.

I is also=12x2+1212ln(x2+1)+C
=12{x2ln(x2+1)}+C1, where C1=C+12

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