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

The integral of (3x+1)/((3x^2+2x+1)^6) is -1/(10(3x^2+2x+1)^5)+C

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

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

- \frac{1}{10 ( 3 x ^{2} + 2 x + 1 ) ^{5}} + C

Solution steps

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

Apply u-substitution

: \int \frac{243 u}{( u ^{2} + 2 ) ^{6}} d u

= \int \frac{243 u}{( u ^{2} + 2 ) ^{6}} d u

Take the constant out:

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

= 243 \cdot \int \frac{u}{( u ^{2} + 2 ) ^{6}} d u

Apply u-substitution

: \int \frac{1}{2 ( v + 2 ) ^{6}} d v

= 243 \cdot \int \frac{1}{2 ( v + 2 ) ^{6}} d v

Take the constant out:

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

= 243 \cdot \frac{1}{2} \cdot \int \frac{1}{( v + 2 ) ^{6}} d v

Apply u-substitution

: \int \frac{1}{w ^{6}} d w

= 243 \cdot \frac{1}{2} \cdot \int \frac{1}{w ^{6}} d w

Apply the Power Rule

: - \frac{1}{5 w ^{5}}

= 243 \cdot \frac{1}{2} (- \frac{1}{5 w ^{5}})

Substitute back

= 243 \cdot \frac{1}{2} - \frac{1}{5 ( ( 3 x + 1 ) ^{2} + 2 ) ^{5}}

Simplify

243 \cdot \frac{1}{2} - \frac{1}{5 ( ( 3 x + 1 ) ^{2} + 2 ) ^{5}} : - \frac{1}{10 ( 3 x ^{2} + 2 x + 1 ) ^{5}}

= - \frac{1}{10 ( 3 x ^{2} + 2 x + 1 ) ^{5}}

Add a constant to the solution

= - \frac{1}{10 ( 3 x ^{2} + 2 x + 1 ) ^{5}} + C

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What is the integral of (3x+1)/((3x^2+2x+1)^6) ?
The integral of (3x+1)/((3x^2+2x+1)^6) is -1/(10(3x^2+2x+1)^5)+C