What is the integral of (2x^2+6sqrt(x)-5x)/(3x^2) ?

The integral of (2x^2+6sqrt(x)-5x)/(3x^2) is 1/3 (2x-(12)/(sqrt(x))-5ln|x|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^2+6sqrt(x)-5x)/(3x^2)

\int \frac{2 x ^{2} + 6 x - 5 x}{3 x ^{2}} d x

\frac{1}{3} (2 x - \frac{12}{x} - 5 ln ∣ x ∣) + C

Solution steps

\int \frac{2 x ^{2} + 6 x - 5 x}{3 x ^{2}} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{2 x ^{2} + 6 x - 5 x}{x ^{2}} d x

Expand

\frac{2 x ^{2} + 6 x - 5 x}{x ^{2}} : 2 + \frac{6}{x ^{\frac{3}{2}}} - \frac{5}{x}

= \frac{1}{3} \cdot \int 2 + \frac{6}{x ^{\frac{3}{2}}} - \frac{5}{x} d x

Apply the Sum Rule:

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

= \frac{1}{3} (\int 2 d x + \int \frac{6}{x ^{\frac{3}{2}}} d x - \int \frac{5}{x} d x)

\int 2 d x = 2 x

\int \frac{6}{x ^{\frac{3}{2}}} d x = - \frac{12}{x}

\int \frac{5}{x} d x = 5 ln ∣ x ∣

= \frac{1}{3} (2 x - \frac{12}{x} - 5 ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{3} (2 x - \frac{12}{x} - 5 ln ∣ x ∣) + C

y = - csc (x) - cos (x)

2 y^{^{''}} + 8 y = 0

\int_{- \infty}^{- 1} e^{2 x} d x

\int 6 x^{2} - 3 x^{3} d x

x + y = 7, x = 0, y = 0

What is the integral of (2x^2+6sqrt(x)-5x)/(3x^2) ?
The integral of (2x^2+6sqrt(x)-5x)/(3x^2) is 1/3 (2x-(12)/(sqrt(x))-5ln|x|)+C