<s>[INST] evaluate the integral $\int x^3(5x^2+3)^5 dx$ How to evaluate for the integral:
$$\int x^3(5x^2+3)^5 dx$$
What I've tried:
For example,
(1) $u = x^2$ whence $du = 2x$ $dx$
(2)$u = ax^2+b$ whence $du = 2ax$ $dx$ 
Therefore,
(3) $u = 5x^2+3$ whence $du = 10x$ $dx$
$$I = \int \frac{du}{10}u^5 = \frac{1}{10} \int \frac{u^6}{6} du= \frac{1}{10}\frac{1}{6}u^6$$
However, this does not lead me to the answer:
$$\frac{(10x^2-1)(5x^2+3)^6}{700}$$
Where did I go wrong and how do I correct this?
 [/INST] There are a few small mistakes in your reasoning. First you forgot the $dx$. Your initial expression should be
$$I=\int x^3(5x^2+3)^5\ dx$$
Applying the change of variables
$$u=5x^2+3 \qquad du=10xdx$$
Then
$$I=\int x^3(5x^2+3)^5\ dx=\frac1{10}\int x^2(5x^2+3)^5\ 10xdx$$
We will need to get rid of $x^2$ as well.
$$u=5x^2+3 \implies x^2=\frac{u-3}{5}$$
Now we are able to proceed
$$I=\frac1{10}\int \frac{u-3}{5}u^5\ du$$
Can you take it from here?
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