What is the integral of 1/(x^2(4+3x)) ?

The integral of 1/(x^2(4+3x)) is 3(1/16 ln|4+3x|-1/16 ln(3|x|)-1/(12x))+C

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integral of 1/(x^2(4+3x))

\int \frac{1}{x ^{2} ( 4 + 3 x )} d x

3 (\frac{1}{16} ln ∣ 4 + 3 x ∣ - \frac{1}{16} ln (3 ∣ x ∣) - \frac{1}{12 x}) + C

Solution steps

\int \frac{1}{x ^{2} ( 4 + 3 x )} d x

Apply u-substitution

: \int \frac{3}{u ( u - 4 ) ^{2}} d u

= \int \frac{3}{u ( u - 4 ) ^{2}} d u

Take the constant out:

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

= 3 \cdot \int \frac{1}{u ( u - 4 ) ^{2}} d u

Take the partial fraction of

\frac{1}{u ( u - 4 ) ^{2}} : \frac{1}{16 u} - \frac{1}{16 ( u - 4 )} + \frac{1}{4 ( u - 4 ) ^{2}}

= 3 \cdot \int \frac{1}{16 u} - \frac{1}{16 ( u - 4 )} + \frac{1}{4 ( u - 4 ) ^{2}} d u

Apply the Sum Rule:

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

= 3 (\int \frac{1}{16 u} d u - \int \frac{1}{16 ( u - 4 )} d u + \int \frac{1}{4 ( u - 4 ) ^{2}} d u)

\int \frac{1}{16 u} d u = \frac{1}{16} ln ∣ u ∣

\int \frac{1}{16 ( u - 4 )} d u = \frac{1}{16} ln ∣ u - 4 ∣

\int \frac{1}{4 ( u - 4 ) ^{2}} d u = - \frac{1}{4 ( u - 4 )}

= 3 (\frac{1}{16} ln ∣ u ∣ - \frac{1}{16} ln ∣ u - 4 ∣ - \frac{1}{4 ( u - 4 )})

Substitute back

u = 4 + 3 x

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

Simplify

3 (\frac{1}{16} ln ∣ 4 + 3 x ∣ - \frac{1}{16} ln ∣ 4 + 3 x - 4 ∣ - \frac{1}{4 ( 4 + 3 x - 4 )}) : 3 (\frac{1}{16} ln ∣ 4 + 3 x ∣ - \frac{1}{16} ln (3 ∣ x ∣) - \frac{1}{12 x})

= 3 (\frac{1}{16} ln ∣ 4 + 3 x ∣ - \frac{1}{16} ln (3 ∣ x ∣) - \frac{1}{12 x})

Add a constant to the solution

= 3 (\frac{1}{16} ln ∣ 4 + 3 x ∣ - \frac{1}{16} ln (3 ∣ x ∣) - \frac{1}{12 x}) + C

y = 7 x^{2}, y = x^{2} + 3

f (x) = \frac{10 x}{x ^{2} - 6}, at x = 4

\int \frac{6 x ^{4} + 24 x ^{2} + 4}{x ^{3} + 4 x} d x

f (x) = 2 x + 3

What is the integral of 1/(x^2(4+3x)) ?
The integral of 1/(x^2(4+3x)) is 3(1/16 ln|4+3x|-1/16 ln(3|x|)-1/(12x))+C