evaluate the integral $\int x^3(5x^2+3)^5 dx$

Question

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?

Answer 1

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?

Answer 2

One possible alternative way to approach this problem consists in using the integration by parts method. Indeed, one has

\begin{align*} \int x^{3}(5x^{2}+3)^{5}\mathrm{d}x = \frac{x^{2}(5x^{2}+3)^{6}}{60} - \frac{1}{30}\int x(5x^{2}+3)^{6}\mathrm{d}x \end{align*}

where the last integral can be computed through the substitution method. Can you proceed from here?

Answer 3

$$\int x^3(5x^2+3)^5 dx=\frac{1}{5}\int(5x^3+3x-3x)(5x^2+3)^5 dx=$$ $$=\frac{(5x^2+3)^7}{5\cdot7\cdot10}-\frac{3(5x^2+3)^6}{5\cdot6\cdot10}+C.$$