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

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

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

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

\frac{1}{112} ln ∣ x - 3 ∣ - \frac{4}{105} ln ∣ 2 x + 1 ∣ + \frac{7}{480} ln 4 x^{2} + 4 x + 16 - \frac{1}{16 15} arctan (\frac{1}{15} (2 x + 1)) + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x - 3 ) ( 2 x + 1 ) ( x ^{2} + x + 4 )} : \frac{1}{112 ( x - 3 )} - \frac{8}{105 ( 2 x + 1 )} + \frac{7 x - 4}{240 ( x ^{2} + x + 4 )}

= \int \frac{1}{112 ( x - 3 )} - \frac{8}{105 ( 2 x + 1 )} + \frac{7 x - 4}{240 ( x ^{2} + x + 4 )} 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{1}{112 ( x - 3 )} d x - \int \frac{8}{105 ( 2 x + 1 )} d x + \int \frac{7 x - 4}{240 ( x ^{2} + x + 4 )} d x

\int \frac{1}{112 ( x - 3 )} d x = \frac{1}{112} ln ∣ x - 3 ∣

\int \frac{8}{105 ( 2 x + 1 )} d x = \frac{4}{105} ln ∣ 2 x + 1 ∣

\int \frac{7 x - 4}{240 ( x ^{2} + x + 4 )} d x = \frac{1}{480 15} (715 ln 4 x^{2} + 4 x + 16 - 30 arctan (\frac{2 x + 1}{15}))

= \frac{1}{112} ln ∣ x - 3 ∣ - \frac{4}{105} ln ∣ 2 x + 1 ∣ + \frac{1}{480 15} (715 ln 4 x^{2} + 4 x + 16 - 30 arctan (\frac{2 x + 1}{15}))

Simplify

\frac{1}{112} ln ∣ x - 3 ∣ - \frac{4}{105} ln ∣ 2 x + 1 ∣ + \frac{1}{480 15} (715 ln 4 x^{2} + 4 x + 16 - 30 arctan (\frac{2 x + 1}{15})) : \frac{1}{112} ln ∣ x - 3 ∣ - \frac{4}{105} ln ∣ 2 x + 1 ∣ + \frac{7}{480} ln 4 x^{2} + 4 x + 16 - \frac{1}{16 15} arctan (\frac{1}{15} (2 x + 1))

= \frac{1}{112} ln ∣ x - 3 ∣ - \frac{4}{105} ln ∣ 2 x + 1 ∣ + \frac{7}{480} ln 4 x^{2} + 4 x + 16 - \frac{1}{16 15} arctan (\frac{1}{15} (2 x + 1))

Add a constant to the solution

= \frac{1}{112} ln ∣ x - 3 ∣ - \frac{4}{105} ln ∣ 2 x + 1 ∣ + \frac{7}{480} ln 4 x^{2} + 4 x + 16 - \frac{1}{16 15} arctan (\frac{1}{15} (2 x + 1)) + C

(x + 5 x) (x^{\frac{3}{2}} - x)

- 2 x^{4} + 7 x e^{x}

\int (b - ∣ x ∣)^{2} d x

f (x) = - 8 x^{2} - 7, at x = 1

\int ln (5 x + 8) d x

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