Definite integral

Calculus (Integrals, Series)
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jacks
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Definite integral

#1

Post by jacks »

Evaluation of definite Integral \(\displaystyle \int \frac{x^3+x+1}{x^4+x^2+1}dx\) and \(\displaystyle \int \frac{x^2+x}{(e^x+x+1)^2}dx\)
Papapetros Vaggelis
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Re: Definite integral

#2

Post by Papapetros Vaggelis »

Hi Jacks.

Let \(\displaystyle{I=\int \dfrac{x^3+x+1}{x^4+x^2+1}\,\mathrm{d}x}\) .

The integration interval is \(\displaystyle{I'=\mathbb{R}}\) . We have that

\(\displaystyle{\begin{aligned} I&=\int \dfrac{x^3+x+1}{x^4+x^2+1}\,\mathrm{d}x\\&=\int \left(\dfrac{x}{x^4+x^2+1}+\dfrac{x^3+1}{x^4+x^2+1}\right)\,\mathrm{d}x\end{aligned}}\)

where :

\(\displaystyle{\begin{aligned} \int \dfrac{x}{x^4+x^2+1}\,\mathrm{d}x&=\dfrac{1}{2}\,\int \dfrac{2\,x}{(x^2)^2+x^2+1}\,\mathrm{d}x\\&=\dfrac{1}{2}\,\int \dfrac{1}{\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\,\mathrm{d}(x^2)\\&=\dfrac{1}{2}\,\int \dfrac{4}{3+\left(2\,x^2+1\right)^2}\,\mathrm{d}(x^2)\\&=2\,\int \dfrac{1}{\displaystyle{3\,\left[1+\left(\dfrac{2\,x^2+1}{\sqrt{3}}\right)^2\right]}}\,\mathrm{d}(x^2)\\&=\dfrac{1}{\sqrt{3}}\,\int \dfrac{1}{1+\left(\dfrac{2}{\sqrt{3}}\,x^2+\dfrac{1}{\sqrt{3}}\right)^2}\,\mathrm{d}\,\left(\dfrac{2}{\sqrt{3}}\,x^2\right)\\&=\dfrac{1}{\sqrt{3}}\,\arctan\,\dfrac{2\,x^2+1}{\sqrt{3}}+c_1\,,c_1\in\mathbb{R}\end{aligned}}\)

and

\(\displaystyle{\begin{aligned} \int \dfrac{x^3+1}{x^4+x^2+1}\,\mathrm{d}x&=\int \dfrac{\left(x+1\right)\,\left(x^2-x+1\right)}{\left(x^2+1\right)^2-x^2}\,\mathrm{d}x\\&=\int \dfrac{\left(x+1\right)\,\left(x^2-x+1\right)}{\left(x^2-x+1\right)\,\left(x^2+x+1\right)}\,\mathrm{d}x\\&=\int \dfrac{x+1}{x^2+x+1}\,\mathrm{d}x\\&=\dfrac{1}{2}\,\int \dfrac{\left(2\,x+1\right)+1}{x^2+x+1}\,\mathrm{d}x\\&=\dfrac{1}{2}\,\int \left(\dfrac{2\,x+1}{x^2+x+1}+\dfrac{4}{3+\left(2\,x+1\right)^2}\right)\,\mathrm{d}x\\&=\,\int \left(\dfrac{1}{2}\cdot \dfrac{2\,x+1}{x^2+x+1}+\dfrac{2}{3}\cdot \dfrac{1}{1+\left(\dfrac{2\,x+1}{\sqrt{3}}\right)^2}\right)\,\mathrm{d}x\\&=\dfrac{1}{2}\,\ln\,\left(x^2+x+1\right)+\dfrac{1}{\sqrt{3}}\,\arctan\,\dfrac{2\,x+1}{\sqrt{3}}+c_2\,,c_2\in\mathbb{R}\end{aligned}}\)

So

\(\displaystyle{I=\dfrac{1}{\sqrt{3}}\,\arctan\,\dfrac{2\,x^2+1}{\sqrt{3}}+\dfrac{1}{2}\,\ln\,\left(x^2+x+1\right)+\dfrac{1}{\sqrt{3}}\,\arctan\,\dfrac{2\,x+1}{\sqrt{3}}+c\,,c\in\mathbb{R}}\) .

Let \(\displaystyle{J=\int \dfrac{x^2+x}{\left(e^{x}+x+1\right)^2}\,\mathrm{d}x}\) .

The function \(\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}\,\,,f(x)=e^{x}+x+1}\) is continuous and strictly increasing with

\(\displaystyle{\lim_{x\to -\infty}f(x)=-\infty\,\,,\lim_{x\to +\infty}f(x)=+\infty}\), so :

\(\displaystyle{f\,\left(\mathbb{R}\right)=\mathbb{R}}\) and thus the function \(\displaystyle{f}\) has only one root \(\displaystyle{x_0}\) .

The integration interval is \(\displaystyle{J'=\left(-\infty,x_0\right)}\) or \(\displaystyle{J'=\left(x_0,+\infty\right)}\) .

\(\displaystyle{\begin{aligned} J&=\int \dfrac{x^2+x}{\left(e^{x}+x+1\right)^2}\,\mathrm{d}x\\&=\int \dfrac{x\,\left(e^{x}+x+1\right)-x\,e^{x}}{\left(e^{x}+x+1\right)^2}\,\mathrm{d}x\\&=\int \left(\dfrac{x}{e^{x}+x+1}-\dfrac{x\,e^{x}}{\left(e^{x}+x+1\right)^2}\right)\,\mathrm{d}x\end{aligned}}\)

where :

\(\displaystyle{\int \dfrac{x}{e^{x}+x+1}\,\mathrm{d}x=\int\left(1-\dfrac{e^{x}+1}{e^{x}+x+1}\right)\,\mathrm{d}x=x-\ln\,\left(e^{x}+x+1\right)+c_{1}\,,c_1\in\mathbb{R}}\)

and

\(\displaystyle{\begin{aligned} \int \dfrac{x\,e^{x}}{\left(e^{x}+x+1\right)^2}\,\mathrm{d}x&=-\int \dfrac{\left(e^{x}+x+1\right)-\left(x+1\right)\,(e^{x}+1)}{\left(e^{x}+x+1\right)^2}\,\mathrm{d}x\\&=-\int \dfrac{\left(x+1\right)'\,\left(e^{x}+x+1\right)-\left(x+1\right)\,\left(e^{x}+x+1\right)'}{\left(e^{x}+x+1\right)^2}\,\mathrm{d}x\\&=-\dfrac{x+1}{e^{x}+x+1}+c_2\,,c_2\in\mathbb{R}\end{aligned}}\)

Finally,

\(\displaystyle{J=x-\ln\,\left|e^{x}+x+1\right|+\dfrac{x+1}{e^{x}+x+1}+c\,,c\in\mathbb{R}}\) .
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Tolaso J Kos
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Re: Definite integral

#3

Post by Tolaso J Kos »

Good evening to both of you... (slightly different)

$$\begin{aligned}
\int \frac{x^3+x+1}{x^4+x^2+1}\,dx &=\int \left ( \frac{1}{2}\frac{2x+1}{x^2+x+1}+\frac{1}{2}\frac{1}{x^2-x+1} \right )\,dx \\
&=\frac{1}{2}\int \frac{2x+1}{x^2+x+1}\,dx+\frac{1}{2}\int \frac{dx}{x^2-x+1} \\
&= \frac{1}{2}\ln \left ( x^2+x+1 \right )+\frac{1}{2}\int \frac{dx}{\left ( x-1/2 \right )^2+\frac{3}{4}}\\
&= \frac{1}{2}\ln \left ( x^2+x+1 \right )+\frac{1}{2}\int \frac{dt}{t^2+3/4}\\
&= \frac{1}{2}\ln \left ( x^2+x+1 \right )+\frac{1}{\sqrt{3}}\tan^{-1}\left ( \frac{2t}{\sqrt{3}} \right )\\
&=\frac{1}{2}\ln \left ( x^2+x+1 \right ) +\frac{1}{\sqrt{3}}\tan^{-1}\left ( \frac{2x-1}{\sqrt{3}} \right ) +c, \;\; c \in \mathbb{R}
\end{aligned}$$

Let me pose a challenge: Prove that the primitives found by me and Vaggelis are equal.
Imagination is much more important than knowledge.
Papapetros Vaggelis
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Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Definite integral

#4

Post by Papapetros Vaggelis »

1st part :

Let \(\displaystyle{A=\left\{\left(x,y\right)\in\mathbb{R}^2: x\,y\neq 1\right\}}\) . Obviously, \(\displaystyle{A\neq \varnothing}\)

casue \(\displaystyle{(0,0)\in A}\) . Consider \(\displaystyle{\left(x,y\right)\in A-\left\{(0,0)\right\}}\) .

(If \(\displaystyle{\left(x,y\right)=\left(0,0\right)}\), then : \(\displaystyle{\arctan\,x+\arctan\,y=0=\arctan\,\left(\frac{x+y}{1-x\,y}\right)}\) ) .

Since the \(\displaystyle{\rm{\tan}}\) - function is

onto \(\displaystyle{\mathbb{R}}\), we have that \(\displaystyle{x=\tan\,t\,\,,y=\tan\,u}\) for some

\(\displaystyle{t\,,u\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)-\left\{0\right\}}\)

and we get \(\displaystyle{-\pi<t+u<\pi}\) . If \(\displaystyle{t+u=\frac{\pi}{2}}\) or \(\displaystyle{t+u=-\frac{\pi}{2}}\) , then :

\(\displaystyle{\tan\,u=\frac{1}{\tan\,t}\implies \tan\,t\,\tan\,u=1\implies x\,y=1}\) , a contradiction. So,

if \(\displaystyle{t+u\in\left(-\pi,-\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right)}\), then :

\(\displaystyle{\tan\,(t+u)=\dfrac{\tan\,t+\tan\,u}{1-\tan\,t\,\tan\,u}=\frac{x+y}{1-x\,y}\implies}\)

\(\displaystyle{\implies t+u=\arctan\,\left(\dfrac{x+y}{1-x\,y}\right)\implies \arctan\,x+\arctan\,y=\arctan\,\left(\dfrac{x+y}{1-x\,y}\right)}\) .

Consider now \(\displaystyle{B=\left\{\left(x,\frac{1}{x}\right)\in\mathbb{R}^2: x\neq 0\right\}=\mathbb{R}^2-A}\) .

\(\displaystyle{\forall\,x\neq 0: \left[\frac{\mathrm{d}}{\mathrm{d}t}\,\arctan\,\frac{1}{t}\right]_{t=x}=-\frac{1}{1+x^2}=\left[\frac{\mathrm{d}}{\mathrm{d}t}\,(-\arctan\,t)\right]_{t=x}}\) ,

so :

\(\displaystyle{\exists\,(c_1\,,c_2\in\mathbb{R}) : \arctan\,\frac{1}{x}=\begin{cases}
-\arctan\,x+c_1\,\,,x>0\\
-\arctan\,x+c_2\,\,,x<0
\end{cases}}\)

and :

\(\displaystyle{\arctan\,1=-\arctan\,1+c_1\iff c_1=2\,\arctan\,1=\frac{\pi}{2}}\)

\(\displaystyle{=\arctan\,(-1)=-\arctan\,(-1)+c_2\iff c_2=2\,\arctan\,(-1)=-\frac{\pi}{2}}\)

which means that \(\displaystyle{\arctan\,\frac{1}{x}=\begin{cases}
-\arctan\,x+\frac{\pi}{2}\,\,,x>0\\
-\arctan\,x-\frac{\pi}{2} \,\,,x<0
\end{cases}}\)

Also, by following the same strategy, we get \(\displaystyle{\arctan\,(-x)=-\arctan\,x\,,x\in\mathbb{R}}\)

2nd part: Let now \(\displaystyle{x\in\mathbb{R}}\) .

\(\displaystyle{\frac{2\,x^2+1}{\sqrt{3}}\cdot \frac{2\,x+1}{\sqrt{3}}=\frac{4\,x^3+2\,x^2+2\,x+1}{3}}\) and

\(\displaystyle{\begin{aligned} \frac{2\,x^2+1}{\sqrt{3}}\cdot \frac{2\,x+1}{\sqrt{3}}=1&\iff 4\,x^3+2\,x^2+2\,x+1=3\\&\iff 2\,x^3+x^2+x-1=0\\&\iff \left(x-\frac{1}{2}\right)\,\left(2\,x^2+2\,x+2\right)=0\\&\iff \left(2\,x-1\right)\,\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]=0\\&\iff x=\frac{1}{2}\end{aligned}}\)

Therefore, if \(\displaystyle{x\neq \frac{1}{2}}\), then :

\(\displaystyle{\begin{aligned} \frac{1}{\sqrt{3}}\,\arctan\,\frac{2\,x^2+1}{\sqrt{3}}+\frac{1}{\sqrt{3}}\,\arctan\,\frac{2\,x+1}{\sqrt{3}}&=\frac{1}{\sqrt{3}}\,\left[\arctan\,\frac{2\,x^2+1}{\sqrt{3}}+\arctan\,\frac{2\,x+1}{\sqrt{3}}\right]\\&=\frac{1}{\sqrt{3}}\,\arctan\,\left(\frac{\displaystyle{\frac{2\,x^2+1}{\sqrt{3}}+\frac{2\,x+1}{\sqrt{3}}}}{1-\displaystyle{\frac{2\,x^2+1}{\sqrt{3}}\cdot \frac{2\,x+1}{\sqrt{3}}}}\right)\\&=\frac{1}{\sqrt{3}}\,\arctan\,\left[\frac{3\,\left(2\,x^2+2\,x+2\right)}{\sqrt{3}\,\left[3-\left(2\,x^2+1\right)\,\left(2\,x+1\right)\right]}\right)\\&=\frac{1}{\sqrt{3}}\,\arctan\,\left[-\frac{\sqrt{3}\,\left(2\,x^2+2\,x+2\right)}{2\,\left(x-\frac{1}{2}\right)\,\left(2\,x^2+2\,x+2\right)}\right]\\&=\frac{1}{\sqrt{3}}\,\arctan\,\left(-\frac{\sqrt{3}}{2\,x-1}\right)\\&=-\frac{1}{\sqrt{3}}\,\arctan\,\left(\frac{\sqrt{3}}{2\,x-1}\right)\\&=\begin{cases}
\frac{1}{\sqrt{3}}\,\arctan\,\left(\frac{2\,x-1}{\sqrt{3}}\right)+\frac{\pi}{2\,\sqrt{3}}\,\,,x<\frac{1}{2}\\
\frac{1}{\sqrt{3}}\,\arctan\,\left(\frac{2\,x-1}{\sqrt{3}}\right)-\frac{\pi}{2\,\sqrt{3}}\,\,,x>\frac{1}{2}
\end{cases}\end{aligned}}\)

and if \(\displaystyle{x=\frac{1}{2}}\), then :

\(\displaystyle{\arctan\,\frac{2\,x^2+1}{\sqrt{3}}+\arctan\,\frac{2\,x+1}{\sqrt{3}}=\arctan\,\frac{\sqrt{3}}{2}+\arctan\,\frac{2}{\sqrt{3}}=\frac{\pi}{2}}\) , whereas :

\(\displaystyle{\arctan\,\frac{2\,x-1}{\sqrt{3}}=0}\)


In conclusion, the primitives are equal.
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