shamieh
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$$\int (z^5 + 9z^2) * (z^3 + 1)^{12} \, dz$$
$$u = z^3 + 1$$
$$z^3 = u - 1$$
$$du = 3z^2 $$
$$1/3 = z^2 dz$$
$$\int (z^5 + 9z^2) * (u - 1 + 1)^{12} $$
$$= \int (z^5 + 9z^2) * u^{12}$$
now I'm stuck..
$$u = z^3 + 1$$
$$z^3 = u - 1$$
$$du = 3z^2 $$
$$1/3 = z^2 dz$$
$$\int (z^5 + 9z^2) * (u - 1 + 1)^{12} $$
$$= \int (z^5 + 9z^2) * u^{12}$$
now I'm stuck..