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How to integrate following

$$\int\frac{u^3}{(u^2+1)^3}du\,?$$

What I did is here:

Used partial fractions

$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$ After solving I got $A=0, B=0, C=1, D=0, E=-1, F=0$ $$\dfrac{u^3}{(u^2+1)^3}=\dfrac{u}{(u^2+1)^2}-\dfrac{u}{(u^2+1)^3}$$ Substitute $u^2+1=t$, $2u\ du=dt$, $u\ du=dt/2$

$$\int\frac{u^3}{(u^2+1)^3}du=\int \frac{dt/2}{t^2}-\int \frac{dt/2}{t^3}$$ $$=\frac12\dfrac{-1}{t}-\frac{1}{2}\dfrac{-1}{2t^2}$$ $$=-\dfrac{1}{2t}+\dfrac{1}{4t^2}$$ $$=-\dfrac{1}{2(u^2+1)}+\dfrac{1}{4(u^2+1)^2}+c$$

My question: Can I integrate this with suitable substitution? Thank you

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10 Answers 10

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Substitute $u=\tan\theta\implies du=\sec^2\theta \ d\theta$ $$\int \frac{u^3}{(u^2+1)^3}du=\int \frac{\tan^3\theta}{(\tan^2\theta+1)^3}\sec^2\theta\ d \theta$$ $$=\int \frac{\tan^3\theta\sec^2\theta}{\sec^6\theta}\ d\theta$$ $$=\int\sin^3\theta\cos\theta d\theta$$ $$=\int\sin^3\theta\ d(\sin\theta)$$ $$=\frac{\sin^4\theta}{4}+C$$

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  • $\begingroup$ Substitute back for $u$ when you're done. $\endgroup$
    – Integrand
    Jul 20, 2020 at 19:22
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Substitute $v=u^2+1$ so that $du=\frac{1}{2u}dv$ to turn the integral into:

$$\int{\frac{u(v-1)}{2uv^3}} dv$$ $$\frac{1}{2}\int{\frac{v-1}{v^3}} dv$$ $$=\frac{1}{2}(\int{\frac{1}{v^2} dv}-\int{\frac{1}{v^3}} dv)$$ $$=\frac{1}{2}(\frac{1}{2v^2}-\frac{1}{v})+C$$ $$=\frac{1}{4v^2}-\frac{1}{2v}+C$$

Substitute $v=u^2+1$ and simplify to get your answer:

$$\dfrac{1}{4(u^2+1)^2}-\dfrac{1}{2(u^2+1)}+c$$ $$=-\dfrac{2u^2+1}{4\left(u^2+1\right)^2}+C$$

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Substitute $u=\sinh t$ to integrate

\begin{align} & \int \dfrac{u^3}{(u^2+1)^3}du= \int \frac{\sinh^3t}{\cosh^5t}dt\\ =&\int\tanh^3td(\tanh t)=\frac14\tanh^4t+C= \frac{u^4}{4(u^2+1)^2}+C \end{align}

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Hint: $\frac {u^{3}} {(u^{2}+1)^{3}}=u\frac {(u^{2}+1)-1} {(u^{2}+1)^{3}}=\frac u {(u^{2}+1)^{2}}-\frac u {(u^{2}+1)^{3}}$. Split then integral into two parts and use the substitution $x=1+u^{2}$ in both . The answer is $-\frac 1 {2(u^{2}+1)} -\frac 1 {4(u^{2}+1)^{2}}+C$

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    $\begingroup$ i did the same thing. please look at my answer $\endgroup$
    – user805287
    Jul 11, 2020 at 23:57
  • $\begingroup$ @SamulGonjo True, but observing that $u^{3}=u((u^{2}+1)-1)$ simplifies the arguement a lot. $\endgroup$ Jul 12, 2020 at 0:00
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This is one case where you can solve the problem without any integration.

Because of the cube in denominator, assume that $$\int\frac{u^3}{(u^2+1)^3}du=\frac{P_n(u)}{(u^2+1)^2}$$ Differentiate both sides and remove the common denominator to get $$u^3=\left(u^2+1\right) P_n'(u)-4 u P_n(u)$$ Comparing the degrees $n=2$; so, write $P_2(u)=a+ b u +c u^2$ to get $$0=b+2 (c-2 a)u-3 b u^2-(2 c+1) u^3$$ Then, $b=0$, $c=-\frac 12$ and $a=-\frac 14$

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  • $\begingroup$ Very good trick. I don't know what you meant by "because of the cube in the denominator", but I can certainly see how multiplying by $u^2+1$ to get a $[(u^2+1)^2]^2$ in the denominator tells us that this is the derivative of a rational function of the form $P(u)/(u^2+1)^2$ because of the quotient rule. $\endgroup$
    – GDGDJKJ
    Jul 12, 2020 at 20:07
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In fact, substitution makes it very easy. Starting with

$$\int \frac{u^3}{(u^2 + 1)^3}\ du$$

take $v = u^2 + 1$, then $dv = 2u\ du$ so $u\ du = \frac{1}{2} dv$. Then we get

$$\begin{align} \int \frac{u^3}{(u^2 + 1)^3}\ du &= \int \frac{u^2 \cdot u}{(u^2 + 1)^3}\ du\\ &= \int \frac{u^2 \cdot \overbrace{(u\ du)}^{\frac{1}{2}\ dv}}{(u^2 + 1)^3}\\ &= \frac{1}{2} \int \frac{u^2\ dv}{v^3}\end{align}$$

Now since $v = u^2 + 1$, we have $u^2 = v - 1$ and

$$\begin{align} \int \frac{u^3}{(u^2 + 1)^3}\ du &= \frac{1}{2} \int \frac{v - 1}{v^3}\ dv\end{align}$$

which is now easy.

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One alternate method is thus: make the substitution $t=u^2$ which gives $dt = 2u\; du \iff du = dt/2 \sqrt t$. Then

$$\mathcal I := \int \frac{u^3}{(u^2+1)^3} du = \int \frac{t \sqrt t}{(t+1)^3} \cdot \frac{dt}{2 \sqrt t} = \frac 1 2\int\frac{t}{(t+1)^3}dt$$

Now let $w = t+1 \implies dw = dt$. Then

$$\mathcal I = \frac 1 2 \int \frac{w-1}{w^3}dw = \frac 1 2 \left( \int \frac 1 {w^2} dw - \int\frac{1}{w^3} dw \right)$$

I imagine you can finish things up from here.

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You can use the partial fraction decomposition $$\dfrac{x^3}{(x^2+1)^3}=\dfrac{3i}{16}\left(\dfrac{1}{(x+i)^2}-\dfrac{1}{(x-i)^2}\right)+\dfrac{1}{8}\left(\dfrac{1}{(x+i)^3}+\dfrac{1}{(x-i)^3}\right)$$ to integrate this function. Then simplify the solution to get rid of complex unit $i.$

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Letting $v = u^2+1$, $dv = 2u\, du, u^2 = v-1$ so

$\begin{array}\\ \int\dfrac{u^3}{(u^2+1)^3}du &=\int\dfrac{(v-1)}{2v^3}dv\\ &=\frac12\int(v^{-2}-v^{-3}dv\\ &=\frac12\left(\dfrac{v^{-1}}{-1}-\dfrac{v^{-2}}{-2}\right)\\ &=\frac12\left(-v^{-1}+\frac12 v^{-2}\right)\\ &=\frac12\left(-\dfrac1{u^2+1}+\dfrac1{2(u^2+1)^2}\right)\\ &=\frac12\dfrac{-2(u^2+1)+1}{2(u^2+1)^2}\\ &=\dfrac{-2u^2-1}{4(u^2+1)^2}\\ &=-\dfrac{2u^2+1}{4(u^2+1)^2}\\ \end{array} $

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I would just attack it with integration by parts (IBP).

The integrand is $\frac{u^3}{(u^2+1)^3}$. You can rewrite this as $u^2 \cdot \frac{u}{(u^2+1)^3}$. The rationale for doing so is to split it up into two terms. The "mental test" you need to apply to successfully IBP is can you 1) differentiate the first term, 2) integrate the second, 3) multiply the two to get something you can integrate again easily? If so, IBP will work well.

Here, applying steps 1 and 2, you end up getting something of the form (neglecting constant multipliers): $u \cdot (u^2+1)^{-2}$, which is easy to integrate again (step 3) so IBP will work fine.

Note that I heavily relied on instant recognition of the "chain rule" form $g'(x)\cdot f(g(x))$, which is a useful skill to pick up early. Otherwise you'll always be stuck with trying substitutions until one works.

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