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

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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1 Answer 1

up vote 0 down vote accepted

we have \begin{align*} \int_0^1 \sqrt{5 + (2t)^2}\, dt &= \frac 12 \int_0^2 \sqrt{5 + x^2}\, dx\\ &= \frac 14\left[x\sqrt{5 + x^2} + 5\log\bigl(x + \sqrt{5 + x^2}\bigr)\right]_0^2\\ &= \frac 14\left(6 + 5\log 5 - 0 - 5 \log \sqrt 5\right)\\ &= \frac 32 + \frac 58 \, \log 5. \end{align*} I don't know what you did wrong ;-), because I don't know what you did at all,

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I think you may need to review the last equality. –  user21436 Mar 25 '12 at 10:23
    
@KannappanSampath Thanks. It should be a $\log$. –  martini Mar 25 '12 at 10:32
    
hi martini, thanks for your answer breakdown. This is what I been looking for. I never posted my workout answer as I'm still new to latex. I think i may get the wrong answer because i never put 1/2 of integral from 0 to 2. Can you explained why this is necessary as opposed to just start from integral 0 to 1 instead? –  kypronite Mar 25 '12 at 10:54
    
@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
    
ah ha, that made sense.thanks once again for you quick response.You saved me from lot of hairpulling :D –  kypronite Mar 25 '12 at 11:08

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