What is the integral of (4x^2+5)/(0.4x^3+1.5x) ?

The integral of (4x^2+5)/(0.4x^3+1.5x) is 3.33333…ln|x|+3.33333…ln|0.4x^2+1.5|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x^2+5)/(0.4x^3+1.5x)

\int \frac{4 x ^{2} + 5}{0.4 x ^{3} + 1.5 x} d x

3.33333 \dots ln ∣ x ∣ + 3.33333 \dots ln 0.4 x^{2} + 1.5 + C

Solution steps

\int \frac{4 x ^{2} + 5}{0.4 x ^{3} + 1.5 x} d x

Factor

\frac{4 x ^{2} + 5}{0.4 x ^{3} + 1.5 x}

= \int \frac{4 x ^{2} + 5}{x ( 0.4 x ^{2} + 1.5 )} d x

Take the partial fraction of

\frac{4 x ^{2} + 5}{x ( 0.4 x ^{2} + 1.5 )} : \frac{10}{3 x} + \frac{2.66666 \dots x}{0.4 x ^{2} + 1.5}

= \int \frac{10}{3 x} + \frac{2.66666 \dots x}{0.4 x ^{2} + 1.5} 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{10}{3 x} d x + \int \frac{2.66666 \dots x}{0.4 x ^{2} + 1.5} d x

\int \frac{10}{3 x} d x = \frac{10}{3} ln ∣ x ∣

\int \frac{2.66666 \dots x}{0.4 x ^{2} + 1.5} d x = 3.33333 \dots ln 0.4 x^{2} + 1.5

= \frac{10}{3} ln ∣ x ∣ + 3.33333 \dots ln 0.4 x^{2} + 1.5

Simplify

= 3.33333 \dots ln ∣ x ∣ + 3.33333 \dots ln 0.4 x^{2} + 1.5

Add a constant to the solution

= 3.33333 \dots ln ∣ x ∣ + 3.33333 \dots ln 0.4 x^{2} + 1.5 + C

\frac{\partial}{\partial x} (sin (x^{2} - 3 x y))

\int_{2}^{5} \frac{1}{x - 3} d x

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

\int ln (x + x^{2}) d x

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

What is the integral of (4x^2+5)/(0.4x^3+1.5x) ?
The integral of (4x^2+5)/(0.4x^3+1.5x) is 3.33333…ln|x|+3.33333…ln|0.4x^2+1.5|+C