Matematické Fórum

Marian · 24. 04. 2018 09:54

Calculate the value of the following integral

$\large\boldsymbol{\int_{0}^{\infty}\frac{\ln (x)}{x^2+2x+4}\,\mathrm{d}x}.$

stuart clark

↑ Marian: let $I=\int^{\infty}_{0}\frac{\ln(x)}{(x+1)^2+3}dx$

Put $x=\frac{4}{y}$ , Then $dx=-\frac{4}{y^2}dy$ and changing limits

$I=\ln(4)\int^{\infty}_{0}\frac{1}{x^2+2x+4}-I\Rightarrow I=\ln(2)\int^{\infty}_{0}\frac{1}{(x+1)^2+3}dx$

$I=\int^{\infty}_{0}\frac{1}{x^2+2x+4}dx=\frac{\pi \cdot \ln(2)}{3\sqrt{3}}$

Marian · 24. 04. 2018 11:17

↑ stuart clark:

Exactly.

stuart clark · 24. 04. 2018 11:22

I have one indefinite integration question:

$\int\frac{(x-2)}{(7x^2-36x+48)\sqrt{x^2-2x-1}}dx$

Options:

$(a)\;\; -2\tan^{-1}\bigg(\frac{\sqrt{3x^2-6x-12}}{9-3x}\bigg)+\mathcal{C}$

$(b)\;\; -2\tan^{-1}\bigg(\frac{\sqrt{4x^2-6x-12}}{9-3x}\bigg)+\mathcal{C}$

$(c)\;\; -2\tan^{-1}\bigg(\frac{\sqrt{3x^2-6x-10}}{9-3x}\bigg)+\mathcal{C}$

$(d)\;\; -2\tan^{-1}\bigg(\frac{\sqrt{3x^2-6x-1}}{9-3x}\bigg)+\mathcal{C}$

Matematické Fórum

#1 24. 04. 2018 09:54

Improper integral

#2 24. 04. 2018 11:13 — Editoval stuart clark (24. 04. 2018 11:13)

Re: Improper integral

#3 24. 04. 2018 11:17

Re: Improper integral

#4 24. 04. 2018 11:22

Re: Improper integral

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