What is the integral of (7x)/(x^2+14x+40) ?

The integral of (7x)/(x^2+14x+40) is 7(5/3 ln|x+10|-2/3 ln|x+4|)+C

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integral of (7x)/(x^2+14x+40)

\int \frac{7 x}{x ^{2} + 14 x + 40} d x

7 (\frac{5}{3} ln ∣ x + 10 ∣ - \frac{2}{3} ln ∣ x + 4 ∣) + C

Solution steps

\int \frac{7 x}{x ^{2} + 14 x + 40} d x

Take the constant out:

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

= 7 \cdot \int \frac{x}{x ^{2} + 14 x + 40} d x

Complete the square

x^{2} + 14 x + 40 : (x + 7)^{2} - 9

= 7 \cdot \int \frac{x}{( x + 7 ) ^{2} - 9} d x

Apply u-substitution

: \int \frac{u - 7}{u ^{2} - 9} d u

= 7 \cdot \int \frac{u - 7}{u ^{2} - 9} d u

Take the partial fraction of

\frac{u - 7}{u ^{2} - 9} : \frac{5}{3 ( u + 3 )} - \frac{2}{3 ( u - 3 )}

= 7 \cdot \int \frac{5}{3 ( u + 3 )} - \frac{2}{3 ( u - 3 )} 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

= 7 (\int \frac{5}{3 ( u + 3 )} d u - \int \frac{2}{3 ( u - 3 )} d u)

\int \frac{5}{3 ( u + 3 )} d u = \frac{5}{3} ln ∣ u + 3 ∣

\int \frac{2}{3 ( u - 3 )} d u = \frac{2}{3} ln ∣ u - 3 ∣

= 7 (\frac{5}{3} ln ∣ u + 3 ∣ - \frac{2}{3} ln ∣ u - 3 ∣)

Substitute back

u = x + 7

= 7 (\frac{5}{3} ln ∣ x + 7 + 3 ∣ - \frac{2}{3} ln ∣ x + 7 - 3 ∣)

Simplify

7 (\frac{5}{3} ln ∣ x + 7 + 3 ∣ - \frac{2}{3} ln ∣ x + 7 - 3 ∣) : 7 (\frac{5}{3} ln ∣ x + 10 ∣ - \frac{2}{3} ln ∣ x + 4 ∣)

= 7 (\frac{5}{3} ln ∣ x + 10 ∣ - \frac{2}{3} ln ∣ x + 4 ∣)

Add a constant to the solution

= 7 (\frac{5}{3} ln ∣ x + 10 ∣ - \frac{2}{3} ln ∣ x + 4 ∣) + C

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What is the integral of (7x)/(x^2+14x+40) ?
The integral of (7x)/(x^2+14x+40) is 7(5/3 ln|x+10|-2/3 ln|x+4|)+C