What is the integral of ((2x))/((4+3x)^2) ?

The integral of ((2x))/((4+3x)^2) is 2(-x/(3(4+3x))+1/9 ln|4+3x|)+C

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integral of ((2x))/((4+3x)^2)

integral

\frac{( 2 x )}{( 4 + 3 x ) ^{2}}

2 (- \frac{x}{3 ( 4 + 3 x )} + \frac{1}{9} ln ∣ 4 + 3 x ∣) + C

Solution steps

\int \frac{2 x}{( 4 + 3 x ) ^{2}} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{( 4 + 3 x ) ^{2}} d x

Apply Integration By Parts

: - \frac{x}{3 ( 4 + 3 x )} - \int - \frac{1}{3 ( 4 + 3 x )} d x

= 2 (- \frac{x}{3 ( 4 + 3 x )} - \int - \frac{1}{3 ( 4 + 3 x )} d x)

\int - \frac{1}{3 ( 4 + 3 x )} d x = - \frac{1}{9} ln ∣ 4 + 3 x ∣

= 2 (- \frac{x}{3 ( 4 + 3 x )} - (- \frac{1}{9} ln ∣ 4 + 3 x ∣))

Simplify

= 2 (- \frac{x}{3 ( 4 + 3 x )} + \frac{1}{9} ln ∣ 4 + 3 x ∣)

Add a constant to the solution

= 2 (- \frac{x}{3 ( 4 + 3 x )} + \frac{1}{9} ln ∣ 4 + 3 x ∣) + C

\frac{1}{( 4 + x )}

f (x) = \frac{1}{x ^{\frac{1}{2}}}

\frac{x + 1}{3 x + 2}

sin^{6} (x) cos^{4} (x)

\frac{( 1 - cos ( x ) ) ^{2}}{sin ^{2} ( x )}

What is the integral of ((2x))/((4+3x)^2) ?
The integral of ((2x))/((4+3x)^2) is 2(-x/(3(4+3x))+1/9 ln|4+3x|)+C