Thread: please help- integration

find the integral of
f (x) = (1+x^2)ArcTan[ x]

Originally Posted by fair_lady0072002

find the integral of
f (x) = (1+x^2)ArcTan[ x]

You'll have to be VERY careful of the domain here, but you can try

x = tan(y)
dx = sec^2(y) dy

Then 1 + x^2 --> 1 + tan^2(y) = sec^2(y)

Your integral then becomes f(y) = sec^4(y)

-Dan

Originally Posted by fair_lady0072002

find the integral of
f (x) = (1+x^2)ArcTan[ x]

My, where you do get these complicated integration from?

Here is what you do,

INTEGRAL (1+x^2)^2*atan(x)*(1+x^2)^{-1} dx

Let, u=atan(x) then, u'=(1+x^2)^{-1}
Then,
sec^2 u=1-x^2
cos^2 u=x^2
Then,
sec^2 u +2cos^2 u=1+x^2
Thus,
INTEGRAL (sec^2 u +2cos^2 u)^2 u*u' dx
Substitution rule it is composition of outer function,
INTEGRAL (sec^2 u+2cos^2 u)^2 du
Open parantheses, (secant*cosine=1)
INTEGRAL (sec^4 u+4cos^4 u+2) du
And this is definitely solvable

Originally Posted by ThePerfectHacker

My, where you do get these complicated integration from?

Here is what you do,

INTEGRAL (1+x^2)^2*atan(x)*(1+x^2)^{-1} dx

Let, u=atan(x) then, u'=(1+x^2)^{-1}
Then,
sec^2 u=1-x^2
cos^2 u=x^2
Then,
sec^2 u +2cos^2 u=1+x^2
Thus,
INTEGRAL (sec^2 u +2cos^2 u)^2 u*u' dx
Substitution rule it is composition of outer function,
INTEGRAL (sec^2 u+2cos^2 u)^2 du
Open parantheses, (secant*cosine=1)
INTEGRAL (sec^4 u+4cos^4 u+2) du
And this is definitely solvable

That's a neat little trick I'll have to remember...

-Dan

Originally Posted by topsquark

That's a neat little trick I'll have to remember...

-Dan

When you are like me (do not use the dx,dy because their meanings have not been definied) you can only use these tricks to introduce the derivative and hence use substitution rule.

And thank you for the reputation.

Hello, fair_lady0072002!

Find the integral of: .(1 + x²) arctan(x) dx

Let: u= arctan(x) . → . x = tan(u) . → . dx = sec²(u) du

. . Then: .1 + x² .= .1 + tan²(x) .= .sec²(x)

Substitute: . ∫ sec²(u) · u · sec²(u) du . = . ∫ u·sec^4(u) du

. . . Good luck!