What is the integral of (29)/((x+1)(x^2+x)) ?

The integral of (29)/((x+1)(x^2+x)) is 29(-ln|x+1|+1/(x+1)+ln|x|)+C

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integral of (29)/((x+1)(x^2+x))

\int \frac{29}{( x + 1 ) ( x ^{2} + x )} d x

29 (- ln ∣ x + 1 ∣ + \frac{1}{x + 1} + ln ∣ x ∣) + C

Solution steps

\int \frac{29}{( x + 1 ) ( x ^{2} + x )} d x

Take the constant out:

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

= 29 \cdot \int \frac{1}{( x + 1 ) ( x ^{2} + x )} d x

Take the partial fraction of

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

= 29 \cdot \int - \frac{1}{x + 1} - \frac{1}{( x + 1 ) ^{2}} + \frac{1}{x} 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

= 29 (- \int \frac{1}{x + 1} d x - \int \frac{1}{( x + 1 ) ^{2}} d x + \int \frac{1}{x} d x)

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

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

\int \frac{1}{x} d x = ln ∣ x ∣

= 29 (- ln ∣ x + 1 ∣ - (- \frac{1}{x + 1}) + ln ∣ x ∣)

Simplify

= 29 (- ln ∣ x + 1 ∣ + \frac{1}{x + 1} + ln ∣ x ∣)

Add a constant to the solution

= 29 (- ln ∣ x + 1 ∣ + \frac{1}{x + 1} + ln ∣ x ∣) + C

y = \frac{e ^{x}}{9 x ^{2}}

\int \frac{( x ^{2} )}{13 - 4 x + x ^{2}} d x

\frac{d}{d x} (\frac{x + 6}{x ^{3} + x - 1})

y^{^{''}} + 4 y^{^{'}} + 3 y = t e^{- t}

x \to 6 lim (\frac{3 x}{x - 6})

What is the integral of (29)/((x+1)(x^2+x)) ?
The integral of (29)/((x+1)(x^2+x)) is 29(-ln|x+1|+1/(x+1)+ln|x|)+C