Integral

Calculus (Integrals, Series)
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Papapetros Vaggelis
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Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Integral

#1

Post by Papapetros Vaggelis »

Find the indefinite integral

\(\displaystyle{I=\int \frac{x^2+2\,x+1+\left(3\,x+1\right)\,\sqrt{x+\ln x}}{x\,\sqrt{x+\ln x}\,\left(x+\sqrt{x+\ln x}\right)}\,dx}\)
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Integral

#2

Post by Papapetros Vaggelis »

We define \(\displaystyle{f:\left(0,+\infty\right)\longrightarrow \mathbb{R}}\) by \(\displaystyle{f(x)=x+\ln x}\) .

This function is continuous and strictly increasing in \(\displaystyle{\left(0,+\infty\right)}\) , so

\(\displaystyle{f\left(\left(0,+\infty\right)\right)=\left(\lim_{x\to 0}f(x),\lim_{x\to +\infty}f(x)\right)=\mathbb{R}}\) .

Therefore, there is only one \(\displaystyle{x_0>0}\) such that \(\displaystyle{f(x_0)=0}\) and then the

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

We observe that \(\displaystyle{\left(x+\sqrt{x+\ln x}\right)'=1+\frac{1}{2\,\sqrt{x+\ln x}}\cdot \left(1+\frac{1}{x}\right)=\frac{2\,x\,\sqrt{x+\ln x}+\left(x+1\right)}{2\,x\,\sqrt{x+\ln x}}\,\,,x\in J}\)

In this way,

\(\displaystyle{\begin{aligned} I&=\int \frac{2\,x\,\sqrt{x+\ln x}+\left(x+1\right)+\left(x+1\right)\sqrt{x+\ln x}+x^2+x}{x\,\sqrt{x+\ln x}\,\left(x+\sqrt{x+\ln x}\right)}\,dx\\&=\int \frac{2\,x\,\sqrt{x+\ln x}+\left(x+1\right)+\left(x+1\right)\,\left(x+\sqrt{x+\ln x}\right)}{x\,\sqrt{x+\ln x}\left(x+\sqrt{x+\ln x}\right)}\,dx\\&=2\,\int \frac{2\,x\,\sqrt{x+\ln x}+\left(x+1\right)}{2\,x\,\sqrt{x+\ln x}\,\left(x+\sqrt{x+\ln x}\right)}\,dx+\int \frac{x+1}{x\,\sqrt{x+\ln x}}\,dx\\&=2\,\int \frac{\left(x+\sqrt{x+\ln x}\right)'}{x+\sqrt{x+\ln x}}\,dx+\int \frac{1}{\sqrt{x+\ln x}}\cdot \left(1+\frac{1}{x}\right)\,dx\\&=2\,\int \frac{\left(x+\sqrt{x+\ln x}\right)'}{x+\sqrt{x+\ln x}}\,dx+2\,\int \frac{\left(x+\ln x\right)'}{2\,\sqrt{x+\ln x}}\,dx\\&=2\,\ln \left(x+\sqrt{x+\ln x}\right)+2\,\sqrt{x+\ln x}+c\,\,,c\in\mathbb{R}\,\,,x\in J\end{aligned}}\)
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