What is the integral of (1-3/y+x) ?

The integral of (1-3/y+x) is y-3ln|y|+xy+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1-3/y+x)

\int (1 - \frac{3}{y} + x) d y

y - 3 ln ∣ y ∣ + x y + C

Solution steps

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

Apply the Sum Rule:

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

= \int 1 d y - \int \frac{3}{y} d y + \int x d y

\int 1 d y = y

\int \frac{3}{y} d y = 3 ln ∣ y ∣

\int x d y = x y

= y - 3 ln ∣ y ∣ + x y

Add a constant to the solution

= y - 3 ln ∣ y ∣ + x y + C

\int_{1}^{4} (x^{3} + \frac{1}{4 x ^{3}}) d x

n = 0 \sum \infty \frac{n ^{6}}{2 ^{n}}

n = 1 \sum \infty - 1^{n}

\frac{d}{d x} (\frac{2 x ^{3} - 7}{x ^{2}})

y = 3 x^{2} - x^{3}, (1, 2)