# can't figure out what I been doing wrong on simple integration question

$$\int_0^1 \sqrt{(\sqrt{5})^2+(2t)^2}\;dt$$

Based on the formula $\int \sqrt{a^2+x^2}\;dx=\frac{1}{2}[x\sqrt{a^2+x^2}+a^2\log(x+\sqrt{a^2+x^2})]$

I just plug in above input into the formula above

However I can only find $3+\frac{5}{2}\log(5)$ but answers that I get from Mathematica is $\frac{3}{2}+\frac{5}{8}\log(5)$

i been trying to figuring out what I been doing wrong for days but I still can't find out what I been doing wrong.

Appreciate if someone can show what I'm been doing wrong

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@KannappanSampath Thanks. It should be a $\log$. –  martini Mar 25 '12 at 10:32
@kypronite As I wanted to apply the formula you stated for $\int \sqrt{a^2 + x^2}\,dx$ I had to get rid of the $2$ in front of $t$ in $\int_0^1 \sqrt{5 + (2t)^2}\, dt$. So I did a substitution $x = 2t$, so $dx = 2\,dt$ and got $\int_0^2 \sqrt{5 + x^2}\, \frac 12\, dx$. –  martini Mar 25 '12 at 10:57