How do I evaluate this indefinite integral?

Question

I am currently working on the following problem:

$\int x (2x+3)^{99}$

I have tried using u-substitution $(u = 2x+3)$ and integration by parts, but have not been able to make any progress that leads me to an answer. I thought about actually computing $(2x+3)^{99}$, but I think that would make the problem even more complicated. How would I go about solving this question?

Answer 1

You have the right idea, but just split the result into $2$ integrations after you make the substitution. In particular,

$$u = 2x + 3 \Rightarrow du = 2dx \Rightarrow dx = \frac{du}{2} \tag{1}\label{eq1}$$

Also,

$$x = \frac{u - 3}{2} \tag{2}\label{eq2}$$

Thus,

$$\int x\left(2x + 3\right)^{99}dx = \int \frac{u - 3}{4}u^{99} du = \frac{1}{4}\int u^{100} du - \frac{3}{4}\int u^{99} du \tag{3}\label{eq3}$$

I assume you should be able to finish the rest.

Answer 2

Integration by parts works: $$\int x(2x+3)^{99}dx=x\cdot\frac{(2x+3)^{100}}{2\cdot100}-\int\frac{(2x+3)^{100}}{2\cdot100}dx$$ Can you end it now?

Answer 3

We have $$\int x(2x+3)^{99} dx=\int\frac{u-3}{4}u^{99}du=\frac{1}{4}\int (u^{100}-3u^{99})du=\frac{1}{4}\left(\frac{u^{101}}{101}-\frac{3u^{100}}{100}\right)+C.$$