What is the integral of (6x-7)/((2x+1)(4x^2+1)) ?

The integral of (6x-7)/((2x+1)(4x^2+1)) is -5/2 ln|2x+1|+5/4 ln|4x^2+1|-arctan(2x)+C

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integral of (6x-7)/((2x+1)(4x^2+1))

\int \frac{6 x - 7}{( 2 x + 1 ) ( 4 x ^{2} + 1 )} d x

- \frac{5}{2} ln ∣ 2 x + 1 ∣ + \frac{5}{4} ln 4 x^{2} + 1 - arctan (2 x) + C

Solution steps

\int \frac{6 x - 7}{( 2 x + 1 ) ( 4 x ^{2} + 1 )} d x

Take the partial fraction of

\frac{6 x - 7}{( 2 x + 1 ) ( 4 x ^{2} + 1 )} : - \frac{5}{2 x + 1} + \frac{10 x - 2}{4 x ^{2} + 1}

= \int - \frac{5}{2 x + 1} + \frac{10 x - 2}{4 x ^{2} + 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

= - \int \frac{5}{2 x + 1} d x + \int \frac{10 x - 2}{4 x ^{2} + 1} d x

\int \frac{5}{2 x + 1} d x = \frac{5}{2} ln ∣ 2 x + 1 ∣

\int \frac{10 x - 2}{4 x ^{2} + 1} d x = \frac{5}{4} ln 4 x^{2} + 1 - arctan (2 x)

= - \frac{5}{2} ln ∣ 2 x + 1 ∣ + \frac{5}{4} ln 4 x^{2} + 1 - arctan (2 x)

Add a constant to the solution

= - \frac{5}{2} ln ∣ 2 x + 1 ∣ + \frac{5}{4} ln 4 x^{2} + 1 - arctan (2 x) + C

s^{2} + 4 s + 13

y^{^{'}} = 3 x + x y

\frac{d}{d x} ((4 x - 6)^{2})

9 x^{2}, \frac{1}{9} x

f (x) = tan (x + 4)

What is the integral of (6x-7)/((2x+1)(4x^2+1)) ?
The integral of (6x-7)/((2x+1)(4x^2+1)) is -5/2 ln|2x+1|+5/4 ln|4x^2+1|-arctan(2x)+C