What is the integral of (4w+3)/(4w^2+6w-1) ?

The integral of (4w+3)/(4w^2+6w-1) is 1/2 ln|16w^2+24w-4|+C

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integral of (4w+3)/(4w^2+6w-1)

integral

\frac{4 w + 3}{4 w ^{2} + 6 w - 1}

\frac{1}{2} ln 16 w^{2} + 24 w - 4 + C

Solution steps

\int \frac{4 w + 3}{4 w ^{2} + 6 w - 1} d w

Apply u-substitution

= \int \frac{u}{u ^{2} - 13} d u

Apply u-substitution

= \int \frac{1}{2 v} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

= \frac{1}{2} ln (4 w + 3)^{2} - 13

Simplify

\frac{1}{2} ln (4 w + 3)^{2} - 13 : \frac{1}{2} ln 16 w^{2} + 24 w - 4

= \frac{1}{2} ln 16 w^{2} + 24 w - 4

Add a constant to the solution

= \frac{1}{2} ln 16 w^{2} + 24 w - 4 + C

What is the integral of (4w+3)/(4w^2+6w-1) ?
The integral of (4w+3)/(4w^2+6w-1) is 1/2 ln|16w^2+24w-4|+C