Evaluate the integral: $\int \frac{x^2}{{(a + bx)}^2} \, dx$

Question

I need help to evaluate the integral of $$\int \frac{x^2}{{(a + bx)}^2} \, \mathrm{d}x$$ Here is my approach so far:

Let $x = \frac{a}{b}\sin(t)$ so $\mathrm{d}x = \frac{a}{b}\cos(t) \, \mathrm{d}t$

Therefore, $$ \int \frac{x^2}{{(a + bx)}^2} \, \mathrm{d}x =\frac{a}{b^3}\int \frac{\sin^2 t}{(1+\sin t)^2} \, \mathrm{d}t$$ Let $u = \sin t$ so $\mathrm{d}u = \cos t \, \mathrm{d}t$ and $$\begin{align*} &\frac{a}{b^3}\int \frac{\sin^2 t}{(1+\sin t)^2} \, \mathrm{d}t\\ &=\frac{a}{b^3}\int \frac{u^2}{(1+u)^2} \, \mathrm{d}u\\ &=\frac{a}{b^3}\left(-2\log|u+1| - \frac{1}{1+u} + u\right)\\ &=\frac{a}{b^3}\left(-2\log|\sin t+1| - \frac{1}{1+\sin t} + \sin t\right)\\ &=\frac{a}{b^3}\left(-2\log\left|1+\frac{bx}{a}\right| - \frac{a}{a+bx} + \frac{bx}{a}\right)\\ &=\frac{a}{b^3}\left(-2\log\left|\frac{a+bx}{a}\right| - \frac{a}{a+bx} + \frac{bx}{a} \right) \end{align*}$$ But $$\frac{1}{b^3} \left[(a+bx) - 2a\log(a+bx) - \frac{a^2}{a+bx} \right] $$ is the given answer.

Can you show me where I made a mistake? If you can, show a different, faster way of solving this integral. Thank you for your help.

Answer 1

One mistake arises here: $$\newcommand{\dd}{\mathrm{d}} \int \frac{x^2}{{(a + bx)}^2} \, \dd x =\frac{a}{b^3}\int \frac{\sin^2 t}{(1+\sin t)^2} \, \dd t$$ Do this calculation more carefully: $$ \int \frac{ \frac{a^2}{b^2} \sin^2 t}{\left( a + b \frac a b \sin t\right)^2 } \, \frac a b \cos t \, \dd t $$ Notice that you forgot the $\cos(t)$ tied to the differential. Factoring out carefully, we get $$ \int \frac{ \frac{a^2}{b^2} \sin^2 t}{\left( a + b \frac a b \sin t\right)^2 } \, \frac a b \cos t \,\dd t = \frac{a^2}{b^2} \frac a b \frac{1}{a^2} \int \frac{ \sin^2 t \cos t}{\left( 1 + \sin t\right)^2 } \,\dd t = \frac{a}{b^3} \int \frac{ \sin^2 t \cos t}{\left( 1 + \sin t\right)^2 } \, \dd t $$ Later on, you end up doing a substitution that yields $\cos(t) \, \dd t = \dd u$, which happens to negate your mistake, but this is purely dumb luck in this case. Be more mindful of replacing the differential in the future.

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Technically speaking you should also include a constant of integration, this being an indefinite integral, but this is a minor oversight.

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Your answer is ultimately equivalent to the claimed answer: any two "correct answers" to an indefinite integral will differ by a constant value, as per the mean value theorem. You can see this in their Desmos graphs:

Your solution in red has the added benefit of being slightly more general and correct, even, since the logarithm should be in absolute values.

To see the equivalence, first notice: $$\begin{align*} &\frac{1}{b^3} \left[(a+bx) - 2a\log(a+bx) - \frac{a^2}{a+bx} \right] \\ &= \frac{a}{b^3} \left[ 1+\frac b a x - 2\log(a+bx) - \frac{a}{a+bx} \right]\\ &= \frac{a}{b^3} \left[ \frac b a x - 2\log(a+bx) - \frac{a}{a+bx} \right] + \frac{a}{b^3}\\ &= \frac{a}{b^3} \left[ \frac b a x - 2 \left( \log(a+bx) - \log(a) \right) - \frac{a}{a+bx} \right] + \frac{a}{b^3} - \frac{2a}{b^3} \log(a) \\ &= \frac{a}{b^3} \left[ \frac b a x - 2 \log \left( \frac{a+bx}{a} \right) - \frac{a}{a+bx} \right] + \underbrace{\frac{a}{b^3} - \frac{2a}{b^3} \log(a)}_{\text{constant!}} \end{align*} $$ The key steps involved are:

• Factor out an $a$ from the brackets
• Add and subtract $- \frac{2a}{b^3} \log(a)$, to help form the quotient in your answer. When added, we can bring that inside the brackets.

Once one actually accounts for the absolute values, you thus see an entirely-equivalent answer.

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To elaborate on another comment: notice that the composition of your substitutions ultimately amounts to letting $$ x = \frac a b w $$ Notice that that's what happens if you apply your substitutions in sequence: $$ x \xrightarrow{x = \frac a b \sin t} \frac a b \sin t \xrightarrow{u = \sin t} \frac a b u $$ Doing this at the outset would give you $$ \frac{a}{b^3} \int \frac{u^2}{ \left( 1 + u \right)^2} \, \dd u $$ with comparatively minimal effort.

Answer 2

$$I=\int \frac{x^2}{{(a + bx)}^2} \,dx$$ Let $a+b x=t$ $$I=\frac 1{b^3}\int \frac{(t-a)^2}{t^2}\,dt=\frac 1{b^3}\int \left( 1-\frac{2 a}{t}+\frac{a^2}{t^2} \right)\, dt$$

Answer 3

$$ \begin{aligned} \int \frac{x^2}{(a+b x)^2} d x=&-\frac{1}{b} \int x^2 d\left(\frac{1}{a+b x}\right) \\ = & -\frac{x^2}{b(a+b x)}+\frac{2}{b} \int \frac{x}{a+b x} d x \\ = & -\frac{x^2}{b(a+b x)}+\frac{2}{b^2} \int \frac{a+b x-a}{a+b x} d x \\ = & -\frac{x^2}{b(a+b x)}+\frac{2}{b^2(a+b x)}-\frac{2 a}{b^2} \ln |a+b x|+C \end{aligned} $$