What is the integral of (x^3+1)/(x^2+7x+12) ?

The integral of (x^3+1)/(x^2+7x+12) is

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integral of (x^3+1)/(x^2+7x+12)

\int \frac{x ^{3} + 1}{x ^{2} + 7 x + 12} d x

\frac{1}{8} (2 x + 7)^{2} - \frac{21}{2} (x + \frac{7}{2}) + 4 (\frac{37}{8} ln 4 x^{2} + 28 x + 48 + \frac{89}{8} ln ∣ 2 x + 8 ∣ - \frac{89}{8} ln ∣ 2 x + 6 ∣) + C

Solution steps

\int \frac{x ^{3} + 1}{x ^{2} + 7 x + 12} d x

Long division

\frac{x ^{3} + 1}{x ^{2} + 7 x + 12} : x - 7 + \frac{37 x + 85}{x ^{2} + 7 x + 12}

= \int x - 7 + \frac{37 x + 85}{x ^{2} + 7 x + 12} d x

Complete the square

x^{2} + 7 x + 12 : (x + \frac{7}{2})^{2} - \frac{1}{4}

= \int x - 7 + \frac{37 x + 85}{( x + \frac{7}{2} ) ^{2} - \frac{1}{4}} d x

Apply u-substitution

: \int u - \frac{7}{2} - 7 + \frac{4 ( 37 ( u - \frac{7}{2} ) + 85 )}{4 u ^{2} - 1} d u

= \int u - \frac{7}{2} - 7 + \frac{4 ( 37 ( u - \frac{7}{2} ) + 85 )}{4 u ^{2} - 1} 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

= \int u d u - \int \frac{7}{2} d u - \int 7 d u + \int \frac{4 ( 37 ( u - \frac{7}{2} ) + 85 )}{4 u ^{2} - 1} d u

\int u d u = \frac{u ^{2}}{2}

\int \frac{7}{2} d u = \frac{7}{2} u

\int 7 d u = 7 u

\int \frac{4 ( 37 ( u - \frac{7}{2} ) + 85 )}{4 u ^{2} - 1} d u = 4 (\frac{37}{8} ln 4 u^{2} - 1 + \frac{89}{8} ln ∣ 2 u + 1 ∣ - \frac{89}{8} ln ∣ 2 u - 1 ∣)

= \frac{u ^{2}}{2} - \frac{7}{2} u - 7 u + 4 (\frac{37}{8} ln 4 u^{2} - 1 + \frac{89}{8} ln ∣ 2 u + 1 ∣ - \frac{89}{8} ln ∣ 2 u - 1 ∣)

Substitute back

u = x + \frac{7}{2}

= \frac{( x + \frac{7}{2} ) ^{2}}{2} - \frac{7}{2} (x + \frac{7}{2}) - 7 (x + \frac{7}{2}) + 4 (\frac{37}{8} ln 4 (x + \frac{7}{2})^{2} - 1 + \frac{89}{8} ln 2 (x + \frac{7}{2}) + 1 - \frac{89}{8} ln 2 (x + \frac{7}{2}) - 1)

Simplify

\frac{( x + \frac{7}{2} ) ^{2}}{2} - \frac{7}{2} (x + \frac{7}{2}) - 7 (x + \frac{7}{2}) + 4 (\frac{37}{8} ln 4 (x + \frac{7}{2})^{2} - 1 + \frac{89}{8} ln 2 (x + \frac{7}{2}) + 1 - \frac{89}{8} ln 2 (x + \frac{7}{2}) - 1) : \frac{1}{8} (2 x + 7)^{2} - \frac{21}{2} (x + \frac{7}{2}) + 4 (\frac{37}{8} ln 4 x^{2} + 28 x + 48 + \frac{89}{8} ln ∣ 2 x + 8 ∣ - \frac{89}{8} ln ∣ 2 x + 6 ∣)

= \frac{1}{8} (2 x + 7)^{2} - \frac{21}{2} (x + \frac{7}{2}) + 4 (\frac{37}{8} ln 4 x^{2} + 28 x + 48 + \frac{89}{8} ln ∣ 2 x + 8 ∣ - \frac{89}{8} ln ∣ 2 x + 6 ∣)

Add a constant to the solution

= \frac{1}{8} (2 x + 7)^{2} - \frac{21}{2} (x + \frac{7}{2}) + 4 (\frac{37}{8} ln 4 x^{2} + 28 x + 48 + \frac{89}{8} ln ∣ 2 x + 8 ∣ - \frac{89}{8} ln ∣ 2 x + 6 ∣) + C

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What is the integral of (x^3+1)/(x^2+7x+12) ?
The integral of (x^3+1)/(x^2+7x+12) is