What is the integral of (y^2+9)/(y^2+4) ?

The integral of (y^2+9)/(y^2+4) is 5/2 arctan(y/2)+y+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (y^2+9)/(y^2+4)

\int \frac{y ^{2} + 9}{y ^{2} + 4} d y

\frac{5}{2} arctan (\frac{y}{2}) + y + C

Solution steps

\int \frac{y ^{2} + 9}{y ^{2} + 4} d y

\frac{y ^{2} + 9}{y ^{2} + 4} = \frac{5}{y ^{2} + 4} + 1

= \int \frac{5}{y ^{2} + 4} + 1 d y

Apply Integral Substitution

= \int 2 (\frac{5}{4 u ^{2} + 4} + 1) d u

Take the constant out:

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

= 2 \cdot \int \frac{5}{4 u ^{2} + 4} + 1 d u

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{5}{4 u ^{2} + 4} d u + \int 1 d u)

\int \frac{5}{4 u ^{2} + 4} d u = \frac{5}{4} arctan (u)

\int 1 d u = u

= 2 (\frac{5}{4} arctan (u) + u)

Substitute back

u = \frac{y}{2}

= 2 (\frac{5}{4} arctan (\frac{y}{2}) + \frac{y}{2})

Simplify

2 (\frac{5}{4} arctan (\frac{y}{2}) + \frac{y}{2}) : \frac{5}{2} arctan (\frac{y}{2}) + y

= \frac{5}{2} arctan (\frac{y}{2}) + y

Add a constant to the solution

= \frac{5}{2} arctan (\frac{y}{2}) + y + C

\frac{\partial}{\partial x} (e^{x} \cdot ln (y))

sin (x^{2})

\frac{d}{d x} (\frac{3 x - 4}{x + 2})

\int \frac{1}{3 x - 1} d x

\int_{- 4}^{4} 4 - x^{2} d x

What is the integral of (y^2+9)/(y^2+4) ?
The integral of (y^2+9)/(y^2+4) is 5/2 arctan(y/2)+y+C