I am trying to find the integral to 3 x ( x + 1 ) 4

Question

I am trying to find the integral to            3      x              (      x      +      1              )        4

Ada Harrington · Accepted Answer

Of course you can use integration by parts! Your calculations are correct, except that

1 / 2

mysteriously became

3 / 2

in the last step, and that you forgot to add C
Also, when you multiply by something with a minus sign, you need to write it as

x (- \frac{1}{3}) (x + 1)^{- 3}

, not

x - \frac{1}{3} (x + 1)^{- 3}

which means something completely different.
To compare your answer with the one from symbolab, note that

\frac{x}{(x + 1)^{3}} = \frac{x + 1 - 1}{(x + 1)^{3}} = \frac{1}{(x + 1)^{2}} - \frac{1}{(x + 1)^{3}} .

rs450nigglix2 · Accepted Answer

You can use integration by parts without substitution as below:

\begin{aligned} \int \frac{3 x}{(x + 1)^{4}} d x & = - \int x d (\frac{1}{(x + 1)^{3}}) \\ = - \frac{x}{(x + 1)^{3}} + \int \frac{d x}{(x + 1)^{3}} \\ = - \frac{x}{(x + 1)^{3}} - \frac{1}{2 (x + 1)^{2}} + C \\ = - \frac{3 x + 1}{2 (x + 1)^{3}} + C \end{aligned}

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