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

The integral of (p/(2x^2)+pq) is -p/(2x)+pqx+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (p/(2x^2)+pq)

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

- \frac{p}{2 x} + pq x + C

Solution steps

\int \frac{p}{2 x ^{2}} + pq 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

= \int \frac{p}{2 x ^{2}} d x + \int pq d x

\int \frac{p}{2 x ^{2}} d x = - \frac{p}{2 x}

\int pq d x = pq x

= - \frac{p}{2 x} + pq x

Add a constant to the solution

= - \frac{p}{2 x} + pq x + C

\int sin (3 x + 7) d x

\int \frac{e ^{2 x}}{7 - 4 e ^{2 x}} d x

\int \frac{3 x - 1}{( 2 x + 1 ) ( x + 3 )} d x

\int (2 x^{3} + \frac{3}{x ^{2}} + 8 e^{x}) d x

\int \frac{x ^{2} - x}{x} d x

What is the integral of (p/(2x^2)+pq) ?
The integral of (p/(2x^2)+pq) is -p/(2x)+pqx+C