1
$\begingroup$

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

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

1 Answers 1

0

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,

  • 0
    ah ha, that made sense.thanks once again for you quick response.You saved me from lot of hairpulling :D2012-03-25