Question

Integrate ln(1+x^2).

Answer 1

my first thought is integration by parts

Answer 2

I had this on one of my homework problems for calc 2. The answer the professor ultimately gave was different from what I had, and so was what wolfram alpha gave me.
\[\int\limits_{ }^{} \ln \left( 1+x^2 \right)dx\]
I set u=1+x^2 and thus had
\[\int\limits_{ }^{} \ln \left(u \right)du\]
Which can be easily integrated, but when you put u back in it isn't the same as what was on the answer sheet.

Answer 3

that is because du doesn't equal dx there

Answer 4

How do you integrate a natural log function by parts if there's no multiplication involved?

Answer 5

times 1

Answer 6

\[\int\limits \ln(1+x^2) dx=\int\limits 1 \cdot \ln(1+x^2) dx\]

Answer 7

So, lnx=u and x=v?

Answer 8

u=ln(1+x^2)
and
dv=1 dx

Answer 9

just had the entire thing worked out and awaiting confirmation and I accidentally deleted it fml

Answer 10

are you doing okay?

Answer 11

\[u=\ln(1+x^2)\]
\[dv=1dx\]
\[v=x\]
\[du=\frac{ 2x }{ x^2+1 }\]
\[\int\limits_{}^{}(\ln(1+x^2)dx=(\ln(1+x^2)*(x))-\int\limits_{}^{}(x)(\frac{ 2x }{ x^2+1 })dx\]
That last integral turns into a mess.

Answer 12

\[\frac{2x^2}{x^2+1}=\frac{2(x^2+1)-2}{x^2+1}=2-\frac{2}{x^2+1}\]

Answer 13

I see. Let me work this out on paper and I'll get back to you. Thanks for your help and your patience, I appreciate it.

Answer 14

you might need to holler at someone else
i have to go 2 should be easy to integrate
integrating1/(x^2+1) should be from memory that this is arctan(x)

Answer 15

That would explain the 2tan^-1(x) in the answer. Still haven't memorized that.