Quick Integration Question (1 Viewer)

[bcd]

Can somebody look at my solution for this question and point out where I'm going wrong?
The answer is [(square root of 5) - 1 ]

Q: find integral of: X / {sq root(1 + X^2)}

My solution:

let u =X^2
du=2X dx
at X=2, u=4 and at X=0, u=0

Therefore integral becomes (with limits 4 and 0):
{ [0.5 du] / [sq root of (1+u)] }
= 0.5 [ ln {(sq root of u) + (sq root of u+1)} ] (limits are 4 and 0]
= 0.5 ln(2 + sq root of 5)

Thanks
bcd

 

[Mill]

You've integrated incorrectly - it's not a log, it's a backwards chain rule.

You may also find the substitution u = 1 + x^2 easier to work with.

 

[bcd]

Ugh! Can't believe I missed that.
Thanks Mill.