What is the integral of 1/(sqrt(3x^2-3x+7)) ?

The integral of 1/(sqrt(3x^2-3x+7)) is 1/(sqrt(3))ln((|sqrt(3)(2x-1)+2sqrt(3x^2-3x+7)|)/5)+C

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integral of 1/(sqrt(3x^2-3x+7))

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

\frac{1}{3} ln (\frac{3 ( 2 x - 1 ) + 2 3 x ^{2} - 3 x + 7}{5}) + C

Solution steps

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

Complete the square

3 x^{2} - 3 x + 7 : 3 (x - \frac{1}{2})^{2} + \frac{25}{4}

= \int \frac{1}{3 ( x - \frac{1}{2} ) ^{2} + \frac{25}{4}} d x

Apply u-substitution

: \int \frac{2}{12 u ^{2} + 25} d u

= \int \frac{2}{12 u ^{2} + 25} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{12 u ^{2} + 25} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2 3} d v

= 2 \cdot \int \frac{sec ( v )}{2 3} d v

Take the constant out:

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

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

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 2 \cdot \frac{1}{2 3} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 2 \cdot \frac{1}{2 3} ln tan (arctan (\frac{2 3}{5} (x - \frac{1}{2}))) + sec (arctan (\frac{2 3}{5} (x - \frac{1}{2})))

Simplify

2 \cdot \frac{1}{2 3} ln tan (arctan (\frac{2 3}{5} (x - \frac{1}{2}))) + sec (arctan (\frac{2 3}{5} (x - \frac{1}{2}))) : \frac{1}{3} ln (\frac{3 ( 2 x - 1 ) + 2 3 x ^{2} - 3 x + 7}{5})

= \frac{1}{3} ln (\frac{3 ( 2 x - 1 ) + 2 3 x ^{2} - 3 x + 7}{5})

Add a constant to the solution

= \frac{1}{3} ln (\frac{3 ( 2 x - 1 ) + 2 3 x ^{2} - 3 x + 7}{5}) + C

4 - x^{2}

\int \frac{x}{9 x ^{2} + 121} d x

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\int \frac{x e ^{x}}{81 + e ^{x}} d x

What is the integral of 1/(sqrt(3x^2-3x+7)) ?
The integral of 1/(sqrt(3x^2-3x+7)) is 1/(sqrt(3))ln((|sqrt(3)(2x-1)+2sqrt(3x^2-3x+7)|)/5)+C