## omgitsana Group Title find the integral one year ago one year ago

1. omgitsana

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2. omgitsana

i attempted to solve it but i can't get the answer.. which is 24

$\frac{1}{3}\ln(x^3+3x)$

just guessed

5. omgitsana

yeah i got up to there but i can't get the answer :/

6. rishabh.mission

1st take x2=t nd find is dx/dy

7. rishabh.mission

its easy but i dont have pen/pencil let me try give few min.

$\huge{\ln(x^2+3x)}\huge{|_{1}^{3}}={\ln(3^2+3(3))-\ln(1^2+3(1))}$
sorry i forgot$\frac{1}{3}$ on boh sides