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I'm currently learning calc 2 and feel I'm making a very silly, obvious mistake with solving the integral $\int\frac{-5\sqrt[3]{x^2}}3 dx$

I guess I'm making an algebraic mistake somewhere, this is how I went about solving it:

$$\int\frac{-5x^\frac{7}3}3 dx$$

$$\int-5x^\frac{-2}3 dx$$

$$\frac{-5x^\frac{1}3}{\frac13} = -15x^\frac{1}3 + C$$

But the actual answer is $-x^\frac{5}3$ + C

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$\root 3\of {x^2}=(x^2)^{1/3}=x^{2\cdot {1\over 3}}=x^{2/3}$. – David Mitra Aug 11 '14 at 13:30
Also, you can move constants to the outside of the integral e.g. $\int-\frac{5}{3}x\,\mathrm{d}x=-\frac{5}{3}\int x \,\mathrm{d}x$ – Jam Aug 11 '14 at 13:40
up vote 2 down vote accepted

$$\int\frac{-5\sqrt[3]{x^2}}3 dx=\frac{-5}3\int{\sqrt[3]{x^2}} dx=\frac{-5}3 \int{x^\color{red}{\frac23}}=\frac{-5}3\cdot \frac35 x^\frac53+C=-x^\frac53+C$$

You incorrectly evaluated $\sqrt[3]{x^2}$, the correct one is in $\color{red}{red}$.

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$$\int \frac{-5x^{\frac{2}{3}}}{3}dx=-\frac{5}{3} \int x^{\frac{2}{3}} dx=-\frac{5}{3} \cdot \frac{3}{5} x^{\frac{5}{3}}+c=-x^{\frac{5}{3}}+c$$

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