integral 2x^2(x^3+1)^(1/2) dx Having the hardest time with this is there a simple way to do it.
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Originally Posted by Nightasylum integral 2x^2(x^3+1)^(1/2) dx Having the hardest time with this is there a simple way to do it. For $\displaystyle \int 2x^2 \sqrt{x^3+1}\,dx $ let $\displaystyle u = x^3+1$. See how that works?
Originally Posted by Nightasylum integral 2x^2(x^3+1)^(1/2) dx Having the hardest time with this is there a simple way to do it. substitution ... $\displaystyle u = x^3+1$ $\displaystyle du = 3x^2 \, dx$ $\displaystyle \frac{2}{3} \int 3x^2\sqrt{x^3+1} \, dx$ $\displaystyle \frac{2}{3} \int \sqrt{u} \, du$ finish ...
Originally Posted by skeeter substitution ... $\displaystyle u = x^3+1$ $\displaystyle du = 3x^2 \, dx$ $\displaystyle \frac{2}{3} \int 3x^2\sqrt{x^3+1} \, dx$ $\displaystyle \frac{2}{3} \int \sqrt{u} \, du$ finish ... Ill get 2/3 (x^3+1)^3/2 + c But that doesnt work so im a little lost.
Originally Posted by Nightasylum Ill get 2/3 (x^3+1)^3/2 + c But that doesnt work so im a little lost. ... you ignored the constant in front of the integral.
Originally Posted by skeeter ... you ignored the constant in front of the integral. O ok I got it. Thanks a ton .
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