I=∫x2sec−1(x)dx
Let x=secθ. Then, θ=sec−1(x) and dx=secθtanθdθ.
I=∫sec2θ(θ)(secθtanθdθ)=∫θsec3θtanθdθ
Note that ∫sec3θtanθdθ=∫sec2θ.d(secθ)=13sec3θ. Then, we can easily perform integration by parts on I by letting:
u=θ ⇒ du=dθ
dv=sec3θtanθdθ ⇒ v=13sec3θ
So:
I=13θsec3θ−13∫sec3θdθ ⋆
We can solve this integral with integration by parts again, now letting:
u=secθ ⇒ du=secθtanθdθ
dv=sec2θdθ ⇒ v=tanθ
So:
I=13θsec3θ−13(secθtanθ−∫secθtan2θdθ)
Letting tan2θ=sec2θ−1 this becomes:
I=13θsec3θ−13secθtanθ+13∫secθ(sec2θ−1)
I=13θsec3θ−13secθtanθ+13∫sec3θdθ−13∫secθdθ
I=13θsec3θ−13secθtanθ−13ln|secθ+tanθ|+13∫sec3θdθ
Earlier (⋆) we noted that I=13θsec3θ−13∫sec3θdθ, so we can set this to be equivalent to the expression we just found.
13θsec3θ−13∫sec3θdθ=13θsec3θ−13secθtanθ−13ln|secθ+tanθ|+13∫sec3θdθ
Rearranging:
−23∫sec3θdθ=−13secθtanθ−13ln|secθ+tanθ|
Which gives:
−13∫sec3θdθ=−16secθtanθ−16ln|secθ+tanθ|
Plugging this into ⋆ yields:
I=13θsec3θ−16secθtanθ−16ln|secθ+tanθ|
We can now return to our original variable x using x=secθ, tanθ=√sec2θ−1=√x2−1 and θ=sec−1(x):
I=13x3sec−1(x)−16x√x2−1−16ln∣∣x+√x2−1∣∣+C
