What is the integral of 7/(3x^2+x+2) ?

The integral of 7/(3x^2+x+2) is (14)/(sqrt(23))arctan((6x+1)/(sqrt(23)))+C

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integral of 7/(3x^2+x+2)

\int \frac{7}{3 x ^{2} + x + 2} d x

\frac{14}{23} arctan (\frac{6 x + 1}{23}) + C

Solution steps

\int \frac{7}{3 x ^{2} + x + 2} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{3 x ^{2} + x + 2} d x

Complete the square

3 x^{2} + x + 2 : 3 (x + \frac{1}{6})^{2} + \frac{23}{12}

= 7 \cdot \int \frac{1}{3 ( x + \frac{1}{6} ) ^{2} + \frac{23}{12}} d x

Apply u-substitution

: \int \frac{12}{36 u ^{2} + 23} d u

= 7 \cdot \int \frac{12}{36 u ^{2} + 23} d u

Take the constant out:

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

= 7 \cdot 12 \cdot \int \frac{1}{36 u ^{2} + 23} d u

Apply Integral Substitution

= 7 \cdot 12 \cdot \int \frac{1}{6 23 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 7 \cdot 12 \cdot \frac{1}{6 23} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 7 \cdot 12 \cdot \frac{1}{6 23} arctan (v)

Substitute back

= 7 \cdot 12 \cdot \frac{1}{6 23} arctan (\frac{6}{23} (x + \frac{1}{6}))

Simplify

7 \cdot 12 \cdot \frac{1}{6 23} arctan (\frac{6}{23} (x + \frac{1}{6})) : \frac{14}{23} arctan (\frac{6 x + 1}{23})

= \frac{14}{23} arctan (\frac{6 x + 1}{23})

Add a constant to the solution

= \frac{14}{23} arctan (\frac{6 x + 1}{23}) + C

\int 3 t^{3} d t

x y^{2} d x - x^{2} y^{2} d y = 0

\frac{d}{d x} (e^{\frac{( - x ^{2} )}{2}})

\int \frac{1}{16 - t ^{2}} d t

f (x) = ln (1 + x)

What is the integral of 7/(3x^2+x+2) ?
The integral of 7/(3x^2+x+2) is (14)/(sqrt(23))arctan((6x+1)/(sqrt(23)))+C