(x^4+1)^8 4x^3 i know what the answer is , i just want to see how to do it
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Maybe by putting $\displaystyle u=x^4+1$ and $\displaystyle du=4x^3$. So you have $\displaystyle \int u^8 du$.
Originally Posted by rock candy (x^4+1)^8 4x^3 i know what the answer is , i just want to see how to do it $\displaystyle I = \int (x^4+1)^8 \times 4x^3 dx $ $\displaystyle \text{Let } u = x^4 + 1 $ $\displaystyle \text{Then } \frac{du}{dx} = 4x^3 $ $\displaystyle \therefore du = 4x^3 dx $ $\displaystyle \text{Hence } I = \int u^8 du = \frac{u^9}{9} + C = (x^4+1)^9 + C $
thanks for clearing that up ... i should of seen that .. i guess im sleepy
It's actually $\displaystyle \frac{(x^4+1)^9}{9} + C$. You forgot to keep the $\displaystyle 9$ in the denominator when you subbed back in.
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