how to integrate this simple looking integral

Question

I am currently trying to integrate $$\int \frac{u^{24}}{u^{25}+1}du $$ .This question was asked by one of my colleagues. At first sight I thought it was easy as it contained no trigo functions and simple but after doing it for $20$ mins im no where. I think there is some trick which I am missing . Thanks

Answer 1

You can use a simple substitution as follows:

Let $a = u^{25} + 1, \implies \frac{da}{du} =25u^{24}$,

$$du = \frac{da}{25u^{24}}$$

$$ \int\frac{u^{24}}{u^{25} + 1}du =\frac{1}{25} \int\frac{1}{a}da$$

I'm sure you can finish off from here.

Answer 2

Also you can:

substitue $t=u^{25}$ and $dt=25u^{24}du$

$$\int\frac{1}{25(t+1)}dt=\frac{1}{25}\ln|t+1|+\mathcal C=\color{red}{\frac{1}{25}\ln|u^{25}+1|+\mathcal C}$$

Answer 3

$$f(u) = u^{25} + 1$$

$$f'(u) = 25 u^{24}$$

Thence the numerator is nothing but the derivative of the denominator, except for a constant which is not a problem. Indeed you have:

$$\frac{1}{25}\int \frac{f'(u)}{f(u)}\ \text{d}u$$

Which is nothing but the logarithmic derivative: $\frac{f'(u)}{f(u)} = \frac{\text{d}}{\text{d}u}\log(f(u))$ hence the result

$$\frac{1}{25}\log(u^{25} + 1)$$

Plus the famous arbitrary constant C.