What is the integral of (48x)/((x+1)^2(3x+1)) ?

The integral of (48x)/((x+1)^2(3x+1)) is 48(1/4 ln|x+1|-1/(2(x+1))-1/4 ln|3x+1|)+C

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integral of (48x)/((x+1)^2(3x+1))

\int \frac{48 x}{( x + 1 ) ^{2} ( 3 x + 1 )} d x

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

Solution steps

\int \frac{48 x}{( x + 1 ) ^{2} ( 3 x + 1 )} d x

Take the constant out:

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

= 48 \cdot \int \frac{x}{( x + 1 ) ^{2} ( 3 x + 1 )} d x

Take the partial fraction of

\frac{x}{( x + 1 ) ^{2} ( 3 x + 1 )} : \frac{1}{4 ( x + 1 )} + \frac{1}{2 ( x + 1 ) ^{2}} - \frac{3}{4 ( 3 x + 1 )}

= 48 \cdot \int \frac{1}{4 ( x + 1 )} + \frac{1}{2 ( x + 1 ) ^{2}} - \frac{3}{4 ( 3 x + 1 )} 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

= 48 (\int \frac{1}{4 ( x + 1 )} d x + \int \frac{1}{2 ( x + 1 ) ^{2}} d x - \int \frac{3}{4 ( 3 x + 1 )} d x)

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

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

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

= 48 (\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} - \frac{1}{4} ln ∣ 3 x + 1 ∣)

Add a constant to the solution

= 48 (\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} - \frac{1}{4} ln ∣ 3 x + 1 ∣) + C

t^{2} y^{^{''}} - t (t + 2) y^{^{'}} + (t + 2) y = 0

y = 4 x^{2} + 3 x (- 3.27)

x \to 2 lim (\frac{( x + 2 )}{x - 2})

\frac{d}{d y} (\frac{1}{3} (y^{2} + 2)^{\frac{3}{2}})

\int cos (2) d x

What is the integral of (48x)/((x+1)^2(3x+1)) ?
The integral of (48x)/((x+1)^2(3x+1)) is 48(1/4 ln|x+1|-1/(2(x+1))-1/4 ln|3x+1|)+C