What is the integral of (2(x^2+1))/(x(x^2+2)) ?

The integral of (2(x^2+1))/(x(x^2+2)) is 1/2 ln|x^2+2|+ln(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2(x^2+1))/(x(x^2+2))

\int \frac{2 ( x ^{2} + 1 )}{x ( x ^{2} + 2 )} d x

\frac{1}{2} ln x^{2} + 2 + ln (x) + C

Solution steps

\int \frac{2 ( x ^{2} + 1 )}{x ( x ^{2} + 2 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{x ^{2} + 1}{x ( x ^{2} + 2 )} d x

Expand

\frac{x ^{2} + 1}{x ( x ^{2} + 2 )} : \frac{x ^{2}}{x ( x ^{2} + 2 )} + \frac{1}{x ( x ^{2} + 2 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{x ^{2}}{x ( x ^{2} + 2 )} d x + \int \frac{1}{x ( x ^{2} + 2 )} d x)

\int \frac{x ^{2}}{x ( x ^{2} + 2 )} d x = \frac{1}{2} ln x^{2} + 2

\int \frac{1}{x ( x ^{2} + 2 )} d x = - \frac{1}{4} (ln x^{2} + 2 - 2 ln (x))

= 2 (\frac{1}{2} ln x^{2} + 2 - \frac{1}{4} (ln x^{2} + 2 - 2 ln (x)))

Simplify

2 (\frac{1}{2} ln x^{2} + 2 - \frac{1}{4} (ln x^{2} + 2 - 2 ln (x))) : \frac{1}{2} ln x^{2} + 2 + ln (x)

= \frac{1}{2} ln x^{2} + 2 + ln (x)

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 2 + ln (x) + C

x \to 4 - lim (\frac{1}{x ^{2} - 16})

\int \frac{x}{( 1 - x ^{2} )} d x

y = ln (x^{2} - 4 x + 1), (4, 0)

\frac{1}{2 x + 8}

\frac{d}{d x} (x^{5 c o s (x)})

What is the integral of (2(x^2+1))/(x(x^2+2)) ?
The integral of (2(x^2+1))/(x(x^2+2)) is 1/2 ln|x^2+2|+ln(x)+C