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

The integral of (2x-1)/(x^2(x+3)) is 7/9 ln|x|-7/9 ln|x+3|+1/(3x)+C

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

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

\frac{7}{9} ln ∣ x ∣ - \frac{7}{9} ln ∣ x + 3 ∣ + \frac{1}{3 x} + C

Solution steps

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

Expand

\frac{2 x - 1}{x ^{2} ( x + 3 )} : \frac{2 x}{x ^{2} ( x + 3 )} - \frac{1}{x ^{2} ( x + 3 )}

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

\int \frac{2 x}{x ^{2} ( x + 3 )} d x = 2 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣)

\int \frac{1}{x ^{2} ( x + 3 )} d x = - \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣

= 2 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣) - (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣)

Simplify

2 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣) - (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣) : \frac{7}{9} ln ∣ x ∣ - \frac{7}{9} ln ∣ x + 3 ∣ + \frac{1}{3 x}

= \frac{7}{9} ln ∣ x ∣ - \frac{7}{9} ln ∣ x + 3 ∣ + \frac{1}{3 x}

Add a constant to the solution

= \frac{7}{9} ln ∣ x ∣ - \frac{7}{9} ln ∣ x + 3 ∣ + \frac{1}{3 x} + C

x \to 0 lim (\frac{( 2. 7 ^{x} - 1 )}{x})

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\int a x + b d x

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

What is the integral of (2x-1)/(x^2(x+3)) ?
The integral of (2x-1)/(x^2(x+3)) is 7/9 ln|x|-7/9 ln|x+3|+1/(3x)+C