Sunday, April 8, 2018

Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 26

Evaluate 31arctan(1x)dx by using Integration by parts.
If we let u=arctan(1x) and dv=dx, then
du=ddx(1x)1+(1x)2=1x21+1x2=1x2+1dx and v=x

So,

so, 31arctan(1x)dx=uvvdu=xarctan(1x)x(1x2+1)dx=xarctan(1x)+xx2+1dx


To evaluate xx2+1dx, we let u1=x2+1 so du1=2xdx

xx2+1dx=1u1(du12)=12du1u1=12lnu1=12ln(x2+1)


Going back to the first equation,
31arctan(1x)dx=xarctan(1x)+12ln(x2+1)
Evaluating from x=1 to x=3;
31arctan(1x)=3π6π4+ln2

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