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Finding value of $$\int\frac{3x^4+5x^3+1}{x^5(x^4+2x^3+1)^3}dx$$

What i try

Because of denominator has higher power

So I have put $x=1/t$ and $dx=-1/t^2dt$

$$\Longrightarrow I=-\int\frac{(t^4+5t+3)\cdot t^{11}}{(t^4+2t+1)^3}dt$$

But i got struck at that point, did not know how to solve it

Help me please

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  • $\begingroup$ I can't see my way through this but it should be $(t^4+2t+1)^3$. Wow! Wolfram alpha just zaps this. But that's cheating ;) $\endgroup$ Feb 16 '20 at 4:37
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    $\begingroup$ $x^4+2x^3+1=(x+1)(x^3+x^2-x+1)$ $\endgroup$
    – MafPrivate
    Feb 16 '20 at 4:42
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Having cheated, how could we have solved the integral under our own power? We could try to solve $$x^2(3x^4+5x^3+1)=(Ax^2+Bx+c)(x^4+2x^3+1)+(Dx^3+Ex^2+Fx+G)(4x^3+6x^2)$$ The latter cubic factor being $\frac d{dx}(x^4+2x^3+1)$. Then on equating coefficients of like powers of $x$ we have $$\begin{align}3&=A+4D\\ 5&=2A+B+6D+4E\\ 0&=2B+6E+F\\ 0&=2C+6F+4G\\ 1&=A+6G\\ 0&=B\\ 0&=C\end{align}$$ So now we can back-substitute $$\begin{align}F&=-6E\\ G&=-\frac32F=9E\\ A&=1-6G=1-54E\\ D&=\frac34-\frac14A=\frac12+\frac{27}2E\\ 5&=2(1-54E)+0+6\left(\frac12+\frac{27}2E\right)+4E=5-23E\end{align}$$ So we conclude that $E=F=G=B=C=0$, $A=1$, and $D=1/2$. So we can say that $$\begin{align}\int\frac{3x^4+5x^3+1}{x^5(x^4+2x^3+1)^3}dx&=\int\frac{dx}{x^5(x^4+2x^3+1)^2}+\frac12\int\frac{\frac d{dx}(x^4+2x^3+1)}{x^4(x^4+2x^3+1)^3}dx\\ &=\int\frac{dx}{x^5(x^4+2x^3+1)^2}-\frac14\cdot\frac1{x^4(x^4+2x^3+1)^2}+\frac14\int\frac{(-4)dx}{x^5(x^4+2x^3+1)^2}\\ &=-\frac14\cdot\frac1{x^4(x^4+2x^3+1)^2}+C\end{align}$$ So we were only trying to reduce the exponent on $(x^4+2x^3+1)^{\color{red}3}$ but got lucky and the rest of the integrals canceled.

I feel so dirty...

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  • $\begingroup$ You can do it without integration but differentiation. Cheers :-) $\endgroup$ Feb 16 '20 at 5:33
  • $\begingroup$ @ClaudeLeibovici Guessing may work if the integral is easy. In the general case it boils down to something like this after $2$ steps of reduction of the order of the quaritc in the denominator and then factoring out $(x+1)$ and doing partial fractions. Besides, I couldn't permit myself to guess because I had already peeked at the teacher's answer key. $\endgroup$ Feb 16 '20 at 5:41
  • $\begingroup$ I fully agree with you. But it was looking so complex and a try was not a big deal. $\endgroup$ Feb 16 '20 at 5:50
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Without integration at all.

Considering$$\int\frac{3x^4+5x^3+1}{x^5(x^4+2x^3+1)^3}\,dx$$ just assume that it something like $$\frac {P(x)} {x^2(x^4+2x^3+1)^2}$$ Differentiate both sides to get $$\frac{3x^4+5x^3+1}{x^5(x^4+2x^3+1)^3}=\frac{-2 (5 x^4+8 x^3+1) P(x)+(x^5+2 x^4+x)P'(x)}{x^3\left(x^4+2x^3+1\right)^3}$$ that is to say $$3x^4+5x^3+1=x^2\left(-2 (5 x^4+8 x^3+1) P(x)+(x^5+2 x^4+x)P'(x) \right)$$ If $P(x)$ is of degree $n$, $P'(x)$ of degree $(n-1)$ the rhs is of degree $(n+6)$ while the lhs is of degree $4$. So $n=-2$.

Let $P(x)=a+\frac b x +\frac c {x^2}$ and replace to get $$3x^4+5x^3+1=-4 c-3 b x-2 a x^2-20 c x^3-x^4 (18 b+12 c)-x^5 (16 a+11 b)-10 a x^6$$ Comparing the coefficients $a=0$ and $b=0$ and $c=-\frac 14$.

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Thanks so much friends (Got it)

Multiply Numerator and Denominator by $x$

$$I=\int\frac{3x^5+5x^4+x}{(x^6+2x^5+x^2)^3}dx$$

Put $x^6+2x^5+x^2=t$ and $(6x^5+10x^4+2x)dx=dt$

$$I=\frac{1}{2}\int t^{-3}dt=-\frac{1}{4t^2}+C$$

$$I=-\frac{1}{4(x^6+2x^5+x^2)^2}+C$$

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  • $\begingroup$ Very smart solution. +1 I think the question was designed for this answer. $\endgroup$ Feb 17 '20 at 8:45

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