What is the integral of (3-2x)/(5x+7) ?

The integral of (3-2x)/(5x+7) is -2/5 x+29/25 ln|5x+7|+C

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integral of (3-2x)/(5x+7)

\int \frac{3 - 2 x}{5 x + 7} d x

- \frac{2}{5} x + \frac{29}{25} ln ∣ 5 x + 7 ∣ + C

Solution steps

\int \frac{3 - 2 x}{5 x + 7} d x

Expand

\frac{3 - 2 x}{5 x + 7} : \frac{3}{5 x + 7} - \frac{2 x}{5 x + 7}

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{3}{5 x + 7} d x - \int \frac{2 x}{5 x + 7} d x

\int \frac{3}{5 x + 7} d x = \frac{3}{5} ln ∣ 5 x + 7 ∣

\int \frac{2 x}{5 x + 7} d x = 2 (\frac{1}{5} x - \frac{7}{25} ln ∣ 5 x + 7 ∣)

= \frac{3}{5} ln ∣ 5 x + 7 ∣ - 2 (\frac{1}{5} x - \frac{7}{25} ln ∣ 5 x + 7 ∣)

Simplify

\frac{3}{5} ln ∣ 5 x + 7 ∣ - 2 (\frac{1}{5} x - \frac{7}{25} ln ∣ 5 x + 7 ∣) : - \frac{2}{5} x + \frac{29}{25} ln ∣ 5 x + 7 ∣

= - \frac{2}{5} x + \frac{29}{25} ln ∣ 5 x + 7 ∣

Add a constant to the solution

= - \frac{2}{5} x + \frac{29}{25} ln ∣ 5 x + 7 ∣ + C

\int_{1}^{2} 9 r^{4} ln (r) d r

y^{^{'}} + 2 t y = t

\frac{d y}{d x} = \frac{2 x}{3 y}

\frac{cos ( x )}{x ^{8}}

y = 8 x sin (x)

What is the integral of (3-2x)/(5x+7) ?
The integral of (3-2x)/(5x+7) is -2/5 x+29/25 ln|5x+7|+C