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

The integral of (6x+4)/(1+(3x+2)^4) is 1/3 arctan((3x+2)^2)+C

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

\int \frac{6 x + 4}{1 + ( 3 x + 2 ) ^{4}} d x

\frac{1}{3} arctan ((3 x + 2)^{2}) + C

Solution steps

\int \frac{6 x + 4}{1 + ( 3 x + 2 ) ^{4}} d x

Apply u-substitution

: \int \frac{2 u}{3 ( 1 + u ^{4} )} d u

= \int \frac{2 u}{3 ( 1 + u ^{4} )} d u

Take the constant out:

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

= \frac{2}{3} \cdot \int \frac{u}{1 + u ^{4}} d u

Apply u-substitution

: \int \frac{1}{2 ( 1 + v ^{2} )} d v

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

Take the constant out:

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

= \frac{2}{3} \cdot \frac{1}{2} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

= \frac{2}{3} \cdot \frac{1}{2} arctan (v)

Substitute back

= \frac{2}{3} \cdot \frac{1}{2} arctan ((3 x + 2)^{2})

Simplify

\frac{2}{3} \cdot \frac{1}{2} arctan ((3 x + 2)^{2}) : \frac{1}{3} arctan ((3 x + 2)^{2})

= \frac{1}{3} arctan ((3 x + 2)^{2})

Add a constant to the solution

= \frac{1}{3} arctan ((3 x + 2)^{2}) + C

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What is the integral of (6x+4)/(1+(3x+2)^4) ?
The integral of (6x+4)/(1+(3x+2)^4) is 1/3 arctan((3x+2)^2)+C