What is the integral of (3x+9)/(4x^2+x^3) ?

The integral of (3x+9)/(4x^2+x^3) is 3/16 ln|x|-3/16 ln|x+4|-9/(4x)+C

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integral of (3x+9)/(4x^2+x^3)

\int \frac{3 x + 9}{4 x ^{2} + x ^{3}} d x

\frac{3}{16} ln ∣ x ∣ - \frac{3}{16} ln ∣ x + 4 ∣ - \frac{9}{4 x} + C

Solution steps

\int \frac{3 x + 9}{4 x ^{2} + x ^{3}} d x

Expand

\frac{3 x + 9}{4 x ^{2} + x ^{3}} : \frac{3 x}{4 x ^{2} + x ^{3}} + \frac{9}{4 x ^{2} + x ^{3}}

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{3 x}{4 x ^{2} + x ^{3}} d x + \int \frac{9}{4 x ^{2} + x ^{3}} d x

\int \frac{3 x}{4 x ^{2} + x ^{3}} d x = 3 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣)

\int \frac{9}{4 x ^{2} + x ^{3}} d x = 9 (- \frac{1}{16} ln ∣ x ∣ - \frac{1}{4 x} + \frac{1}{16} ln ∣ x + 4 ∣)

= 3 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣) + 9 (- \frac{1}{16} ln ∣ x ∣ - \frac{1}{4 x} + \frac{1}{16} ln ∣ x + 4 ∣)

Simplify

3 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣) + 9 (- \frac{1}{16} ln ∣ x ∣ - \frac{1}{4 x} + \frac{1}{16} ln ∣ x + 4 ∣) : \frac{3}{16} ln ∣ x ∣ - \frac{3}{16} ln ∣ x + 4 ∣ - \frac{9}{4 x}

= \frac{3}{16} ln ∣ x ∣ - \frac{3}{16} ln ∣ x + 4 ∣ - \frac{9}{4 x}

Add a constant to the solution

= \frac{3}{16} ln ∣ x ∣ - \frac{3}{16} ln ∣ x + 4 ∣ - \frac{9}{4 x} + C

What is the integral of (3x+9)/(4x^2+x^3) ?
The integral of (3x+9)/(4x^2+x^3) is 3/16 ln|x|-3/16 ln|x+4|-9/(4x)+C