Evaluate ∫1(x2+1)2dx

Adela Brown

Adela Brown

Answered

2022-01-05

Evaluate 1(x2+1)2dx

Answer & Explanation

nghodlokl

nghodlokl

Expert

2022-01-06Added 33 answers

dx(x2+1)2
Set x=tan(u) and dx=sec2(u)du. Then (x2+1)2=(tan2(u)+1)2=sec4(u) and u=arctan(x)
=cos2(u)du=12cos(2u)du+121du
=u2+14sin(2u)+C
=x2arctan(x)+x+arctan(x)2x2+2+C
Laura Worden

Laura Worden

Expert

2022-01-07Added 45 answers

There is a faster way. Substitute
x=tan(z) dx=sec2(z)dz
thus
(x2+1)2=(tan2(z)+1)2=sec4(z) z=arctan(x)
And remembering that
1sec2=cos2
your integral is simply
cos2(z)dz
Which is trivial and left to you.
The tangent/secant substitution is a great technique which most of people ignore. Study it, and you will solve lots of awesome integrals!
Final result:
x2arctan(x)+x+arctan(x)2x2+2
star233

star233

Expert

2022-01-11Added 238 answers

Let us find d[(ax+b)/(x2+1)n]dx
=ax22bx+a(x2+1)n+1=a(x2+1)2bx+2a(x2+1)n+1=a(x2+1)n+2a2bx(x2+1)n+1
Integrating both sides,
b=02dx(x2+1)n+1=dx(x2+1)n+x(x2+1)nn=12dx(x2+1)2=dxx2+1+x(x2+1)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?