Trig substitution: integrate sqrt(16+x^2) over x

[skyturnred]

Homework Statement

\int\frac{\sqrt{16+x^{2}}}{x}

Homework Equations

The Attempt at a Solution

set x=4tant
dx=4sec^{2}t dt

so after plugging in and using a quick trig identity I get:

\int\frac{16(sec^{2}t)*4sec^{2}t dt}{4tant}

Then after a quick cleanup:

16*\int tan^{-1}t*sec^{4}t *dt

I think I made a mistake, but if not, what next?

 

[LCKurtz]

Homework Statement

\int\frac{\sqrt{16+x^{2}}}{x}

Homework Equations

The Attempt at a Solution

set x=4tant
dx=4sec^{2}t dt

so after plugging in and using a quick trig identity I get:

\int\frac{16(sec^{2}t)*4sec^{2}t dt}{4tant}

Then after a quick cleanup:

16*\int tan^{-1}t*sec^{4}t *dt

I think I made a mistake, but if not, what next?

Best not to use $\tan^{-1}$ for $1/\tan$ since that may be confused with arctan. But write $\sec^4(x)= \sec^2 x(1 + \tan^2 x)$ and try a u - substitution $u=\tan x$.

 

[jimbobian]

I think you've made a mistake, where has the square root gone?

 

[skyturnred]

I definitely made a mistake. I forgot the square root sign, which changes everything. Thanks!