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

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

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

integral

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{3}{9 u ^{2} + 2} d u

= \int \frac{3}{9 u ^{2} + 2} d u

Take the constant out:

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

= 3 \cdot \int \frac{1}{9 u ^{2} + 2} d u

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{sin ^{4} ( x )}{sin ( x )}

(x^{2} - 4)^{\frac{1}{2}}

\frac{1}{( 100 + x ^{2} ) ^{2}}

\frac{- 2}{t}

(5 x - 4)^{3}

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