What is the integral of (10x+2)/((3x+1)(x-1)) ?

The integral of (10x+2)/((3x+1)(x-1)) is 10/3 (ln|30x-30|-4(-(ln|30x+10|)/(40)+(ln|30x-30|)/(40)))+C

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integral of (10x+2)/((3x+1)(x-1))

\int \frac{10 x + 2}{( 3 x + 1 ) ( x - 1 )} d x

\frac{10}{3} (ln ∣ 30 x - 30 ∣ - 4 (- \frac{ln ∣ 30 x + 10 ∣}{40} + \frac{ln ∣ 30 x - 30 ∣}{40})) + C

Solution steps

\int \frac{10 x + 2}{( 3 x + 1 ) ( x - 1 )} d x

Apply u-substitution

: \int \frac{10 u}{( 3 u + 4 ) ( u - 12 )} d u

= \int \frac{10 u}{( 3 u + 4 ) ( u - 12 )} d u

Take the constant out:

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

= 10 \cdot \int \frac{u}{( 3 u + 4 ) ( u - 12 )} d u

Apply u-substitution

: \int \frac{v - 4}{3 v ( v - 40 )} d v

= 10 \cdot \int \frac{v - 4}{3 v ( v - 40 )} d v

Take the constant out:

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

= 10 \cdot \frac{1}{3} \cdot \int \frac{v - 4}{v ( v - 40 )} d v

Expand

\frac{v - 4}{v ( v - 40 )} : \frac{v}{v ( v - 40 )} - \frac{4}{v ( v - 40 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 10 \cdot \frac{1}{3} (\int \frac{v}{v ( v - 40 )} d v - \int \frac{4}{v ( v - 40 )} d v)

\int \frac{v}{v ( v - 40 )} d v = ln ∣ v - 40 ∣

\int \frac{4}{v ( v - 40 )} d v = 4 (- \frac{1}{40} ln ∣ v ∣ + \frac{1}{40} ln ∣ v - 40 ∣)

= 10 \cdot \frac{1}{3} (ln ∣ v - 40 ∣ - 4 (- \frac{1}{40} ln ∣ v ∣ + \frac{1}{40} ln ∣ v - 40 ∣))

Substitute back

= 10 \cdot \frac{1}{3} (ln ∣ 3 (10 x + 2) + 4 - 40 ∣ - 4 (- \frac{1}{40} ln ∣ 3 (10 x + 2) + 4 ∣ + \frac{1}{40} ln ∣ 3 (10 x + 2) + 4 - 40 ∣))

Simplify

10 \cdot \frac{1}{3} (ln ∣ 3 (10 x + 2) + 4 - 40 ∣ - 4 (- \frac{1}{40} ln ∣ 3 (10 x + 2) + 4 ∣ + \frac{1}{40} ln ∣ 3 (10 x + 2) + 4 - 40 ∣)) : \frac{10}{3} (ln ∣ 30 x - 30 ∣ - 4 (- \frac{ln ∣ 30 x + 10 ∣}{40} + \frac{ln ∣ 30 x - 30 ∣}{40}))

= \frac{10}{3} (ln ∣ 30 x - 30 ∣ - 4 (- \frac{ln ∣ 30 x + 10 ∣}{40} + \frac{ln ∣ 30 x - 30 ∣}{40}))

Add a constant to the solution

= \frac{10}{3} (ln ∣ 30 x - 30 ∣ - 4 (- \frac{ln ∣ 30 x + 10 ∣}{40} + \frac{ln ∣ 30 x - 30 ∣}{40})) + C

\int x^{2} + 8 d x

y

9 y^{^{''}} + 6 y^{^{'}} + y = 0

(2 y - x^{2} y) y^{^{'}} = x (y + 2)^{2}

y = \frac{sin ( 2 x ) + 3 cos ( x )}{x + 1}

What is the integral of (10x+2)/((3x+1)(x-1)) ?
The integral of (10x+2)/((3x+1)(x-1)) is 10/3 (ln|30x-30|-4(-(ln|30x+10|)/(40)+(ln|30x-30|)/(40)))+C