Hans Georg Schaathun
Oktober 2016
Finn fylgjande integral
1. $$ \int \frac{1}{x^2+a^2} dx = $$ | 2. $$ \int \frac{1}{x^2-a^2} dx = $$ |
3. $$ \int \frac{x}{x^2+a^2} dx = $$ | 4. $$ \int \frac{x}{x^2-a^2} dx = $$ |
$$ \int \frac{x}{x^2+a^2} dx = $$ | $$u = x^2+a^2a$$ |
$$ \int \frac12\cdot\frac{1}u du = $$ | $$\frac{du}{dx} = 2x$$ |
$$ \frac12\cdot\ln |u|= $$ | |
$$ \frac12\cdot\ln(x^2+a^2) $$ |
$$ \int \frac{x}{x^2-a^2} dx = $$
$$ I = \int \frac{1}{x^2+a^2} dx = $$ | $$ \frac{d}{dx} \tan^{-1} x = \frac1{1+x^2}$$ |
$$ I = \int \frac1{a^2}\cdot\frac{1}{(x/a)^2+1} dx $$ | |
$$ I = \int \frac1{a^2}\cdot\frac{1}{u^2+1} dx $$ | $$u = \frac xa$$ |
$$ I = \int \frac1{a^2}\cdot\frac{1}{u^2+1} adu $$ | $$a\cdot du = dx$$ |
$$ I = \frac1a\int \frac{1}{u^2+1} du $$ | |
$$ I = \frac1a\tan^{-1} u + C = \frac{\tan^{-1} \frac xa}a + C$$ |
$$ \int \frac{1}{x^2-a^2} dx = $$
$$ I = \int \frac{1}{x^2-a^2} dx = \frac{1}{2a}\int \frac1{x-a}-\frac1{x+a}dx $$
$$ I = \frac{1}{2a} \bigg( \int \frac1{x-a}dx-\int\frac1{x+a}dx\bigg) $$
$$ I = \frac{1}{2a} \big( \ln|x-a| + C_1 - \ln|x+a| - C_2\big) $$
$$ I = \frac{1}{2a} \ln\frac{|x-a|}{|x+a|} + C$$