What is the integral of (5(x^2+3))/(x(x^2+5)) ?

The integral of (5(x^2+3))/(x(x^2+5)) is ln|x^2+5|+3ln|x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5(x^2+3))/(x(x^2+5))

\int \frac{5 ( x ^{2} + 3 )}{x ( x ^{2} + 5 )} d x

ln x^{2} + 5 + 3 ln ∣ x ∣ + C

Solution steps

\int \frac{5 ( x ^{2} + 3 )}{x ( x ^{2} + 5 )} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{2} + 3}{x ( x ^{2} + 5 )} d x

Expand

\frac{x ^{2} + 3}{x ( x ^{2} + 5 )} : \frac{x ^{2}}{x ( x ^{2} + 5 )} + \frac{3}{x ( x ^{2} + 5 )}

Apply the Sum Rule:

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

= 5 (\int \frac{x ^{2}}{x ( x ^{2} + 5 )} d x + \int \frac{3}{x ( x ^{2} + 5 )} d x)

\int \frac{x ^{2}}{x ( x ^{2} + 5 )} d x = \frac{1}{2} ln x^{2} + 5

\int \frac{3}{x ( x ^{2} + 5 )} d x = 3 (\frac{1}{5} ln ∣ x ∣ - \frac{1}{10} ln x^{2} + 5)

= 5 (\frac{1}{2} ln x^{2} + 5 + 3 (\frac{1}{5} ln ∣ x ∣ - \frac{1}{10} ln x^{2} + 5))

Simplify

5 (\frac{1}{2} ln x^{2} + 5 + 3 (\frac{1}{5} ln ∣ x ∣ - \frac{1}{10} ln x^{2} + 5)) : ln x^{2} + 5 + 3 ln ∣ x ∣

= ln x^{2} + 5 + 3 ln ∣ x ∣

Add a constant to the solution

= ln x^{2} + 5 + 3 ln ∣ x ∣ + C

(sin (t))^{2}

\int 2 t \cdot (- cos (2 t)) d t

f (x) = e^{x}, at x = 5

x \to 0 lim (\frac{2}{1 - 2 ^{\frac{1}{x}}})

t^{3} \frac{d y}{d t} + 4 t^{2} y = e^{- t}, y (- 1) = 0

What is the integral of (5(x^2+3))/(x(x^2+5)) ?
The integral of (5(x^2+3))/(x(x^2+5)) is ln|x^2+5|+3ln|x|+C