What Is the Correct Approach to Integrate 2*arctan(x) by Parts?

[PCSL]

Homework Statement

problem: \int2arctanx dx
2\intarctan dx

u=arctanx
du=1/(1+x²)
v=x
dv=dx

xarctanx-\intx/(1+x²)

integrate by parts a second time...

u=x
du=dx
v=arctanx
dv=1/1+x²

xarctanx-\intarctanx

My final answer I get it 2xarctanx-2xarctanx+2/x²+1 which is obviously wrong. Thanks.

 

[rock.freak667]

Homework Statement

problem: \int2arctanx dx
2\intarctan dx

u=arctanx
du=1/(1+x²)
v=x
dv=dx

xarctanx-\intx/(1+x²)

Don't go integration by parts here, just do a substitution of u=1+x²

 

[PCSL]

Don't go integration by parts here, just do a substitution of u=1+x²

Thanks, I got the answer. This may seem like a dumb question, but how come integration by parts didn't work for this part?

 

[HallsofIvy]

Homework Statement

problem: \int2arctanx dx
2\intarctan dx

u=arctanx
du=1/(1+x²)
v=x
dv=dx

xarctanx-\intx/(1+x²)

integrate by parts a second time...

u=x
du=dx
v=arctanx
dv=1/1+x²

xarctanx-\intarctanx

My final answer I get it 2xarctanx-2xarctanx+2/x²+1 which is obviously wrong. Thanks.[/QUOTE]
Because 2/(x^ 1) is the <b>derivative</b> of arctan(x), not the integral.<br /> <br /> Your choice of &quot;u&quot; and &quot;dv&quot; are just the results you got from the first integration by parts so you are just reversing the first integration. What you would correctly get is <br /> x arctan(x)- x arctan(x)+ \int arctan(x)dx= \int arctan(x)dx<br /> exactly what you started with.