Before anything else, I would like to apologize for my english as it is not my mother tongue and I may use it unperfectly.

I would like to solve $\int_a^b\frac{1}{t(t+2)}dt$. Of course, it is doable by $\frac{1}{t(t+2)} = \frac{1}{2t} + \frac{1}{2t+2}$ and then integrating. But I was hoping to solve it by using substitution.

I know the formula to be $$ \int_a^b f'(g(x))g'(x)dx = \int_{g(a)}^{g(b)}f(x)dx$$

it seems to me that I can apply the formula here. Let : $$f(x) = ln(x) \qquad f'(x) = \frac1x$$ $$g(x) = \frac{2}{x}+1 \qquad g'(x) = -\frac{2}{x^2}$$

Therefore $$\int_a^b\frac{1}{x(x+2)}dx = \int_a^b\frac{1}{x^2(1+\frac{2}{x})}dx$$ $$ \int_a^b\frac{1}{x(x+2)}dx = -\frac{1}{2}\int f'(g(x))g'(x)dx$$

I then apply the formula : $$ -\frac{1}{2}\int_a^b f'(g(x))g'(x)dx = -\frac{1}{2}\int_{g(a)}^{g(b)}f(x)dx $$

$$\int_a^b\frac{1}{x(x+2)}dx = -\frac{1}{2}\int_{\frac{2}{a}+1}^{\frac{2}{b}+1}ln(x)dx$$

But those two ($\int_a^b\frac{1}{x(x+2)}dx$ and $-\frac{1}{2}\int_{\frac{2}{a}+1}^{\frac{2}{b}+1}ln(x)dx$) gives widly different result for a and b chosen randomly (for instance for (a,b) = (1,2), I get 0.202 = 0.454). Please, could someone be so kind to help me figure out what I did wrong ?

Thank you for your time.

  • $\begingroup$ Partial factions is a little off. Looks like you already integrated but still have integral sign $\endgroup$
    – randomgirl
    Apr 16, 2019 at 18:01
  • $\begingroup$ @randomgirl indeed you're right $\frac{1}{t(t+2)} = \frac{1}{2t} - \frac{1}{2t+2}$. I'll change that in my question. But my point is on computing $\int\frac{1}{x(x+2)}$ and the other directly $\endgroup$
    – NRagot
    Apr 16, 2019 at 18:17

2 Answers 2


It's because the formula should be $$\sf{\int_a^b f'(g(x))g'(x)\,dx = \color{red}{\int_{g(a)}^{g(b)}f'(x)\,dx}=[f(x)]_{g(a)}^{g(b)}=f(g(b))-f(g(a))}$$ not $\sf{\int_{g(a)}^{g(b)} f(x)\,dx}$ on the RHS, as the substitution $\sf{u=g(x)}$ confirms this. Thus you have that $$\sf{\int_a^b\frac1{x(x+2)}\,dx=-\frac12\left[\ln\left(\frac2b+1\right)-\ln\left(\frac2a+1\right)\right]=\frac12\left[\ln\frac b{b+2}-\ln\frac a{a+2}\right].}$$

  • $\begingroup$ thank you ! You are indeed right. It seems that I didn't know my formulas as well as I thought. I'll re-read my learning materials. Again, thank you for your time :-). $\endgroup$
    – NRagot
    Apr 16, 2019 at 18:59
  • $\begingroup$ No problem. I quite like this method, though I would have just gone for partial fractions in an exam. The derivative sign is often very easy to miss out! $\endgroup$
    – TheSimpliFire
    Apr 16, 2019 at 19:00

$\int \frac{1}{t(t+2)} \mathrm{d} t$

This integral could be solved using partial fraction decomposition, but there is a simpler way.

Expand fraction by $\frac{1}{t^{2}}$

$=\int \frac{1}{\left(\frac{2}{t}+1\right) t^{2}} d t$

Substitute $u=\frac{2}{t}+1 \longrightarrow \mathrm{d} t=-\frac{t^{2}}{2} \mathrm{d} u$

$=-\frac{1}{2} \int \frac{1}{u} \mathrm{d} u$

$\begin{aligned} \text { Now solving: } \\ & \int \frac{1}{u} \mathrm{d} u \\ \text { This is a standard integral: } \\ &=\ln (u) \end{aligned}$

$\begin{aligned} \text { Plug in solved integrals: } \\ &-\frac{1}{2} \int \frac{1}{u} \mathrm{d} u \\ &=-\frac{\ln (u)}{2} \end{aligned}$

Undo substitution $u=\frac{2}{t}+1 :$ $=-\frac{\ln \left(\frac{2}{t}+1\right)}{2}$

The problem is solved. Apply the absolute value function to arguments of logarithm functions in order to extend the antiderivative's domain:

$\begin{aligned} & \int \frac{1}{t(t+2)} \mathrm{d} t \\=-& \frac{\ln \left(\left|\frac{2}{t}+1\right|\right)}{2}+C \end{aligned}$

  • $\begingroup$ I'm sure you can do the rest , when computing a given definite integral $\endgroup$ Apr 16, 2019 at 18:37
  • $\begingroup$ I also know of integral-calculator.com ;-). The way the do substitution always left me confused. I'd rather use a proven formula. I don't even know if that is a "correct" way to do so. But thank you anyway for your time. $\endgroup$
    – NRagot
    Apr 16, 2019 at 19:03
  • $\begingroup$ It is correct , I just didnt want to waste time doing it myself. What they did is they used a clever step $\int \frac{1}{\left(\frac{2}{t}+1\right) t^{2}} d t$ $\endgroup$ Apr 16, 2019 at 20:02

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