What is the integral of 1/(2x^2-5x+7) ?

The integral of 1/(2x^2-5x+7) is 2/(sqrt(31))arctan((4x-5)/(sqrt(31)))+C

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integral of 1/(2x^2-5x+7)

\int \frac{1}{2 x ^{2} - 5 x + 7} d x

\frac{2}{31} arctan (\frac{4 x - 5}{31}) + C

Solution steps

\int \frac{1}{2 x ^{2} - 5 x + 7} d x

Complete the square

2 x^{2} - 5 x + 7 : 2 (x - \frac{5}{4})^{2} + \frac{31}{8}

= \int \frac{1}{2 ( x - \frac{5}{4} ) ^{2} + \frac{31}{8}} d x

Apply u-substitution

: \int \frac{8}{16 u ^{2} + 31} d u

= \int \frac{8}{16 u ^{2} + 31} d u

Take the constant out:

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

= 8 \cdot \int \frac{1}{16 u ^{2} + 31} d u

Apply Integral Substitution

= 8 \cdot \int \frac{1}{4 31 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 8 \cdot \frac{1}{4 31} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 8 \cdot \frac{1}{4 31} arctan (v)

Substitute back

= 8 \cdot \frac{1}{4 31} arctan (\frac{4}{31} (x - \frac{5}{4}))

Simplify

8 \cdot \frac{1}{4 31} arctan (\frac{4}{31} (x - \frac{5}{4})) : \frac{2}{31} arctan (\frac{4 x - 5}{31})

= \frac{2}{31} arctan (\frac{4 x - 5}{31})

Add a constant to the solution

= \frac{2}{31} arctan (\frac{4 x - 5}{31}) + C

\int (x + 1) e^{x^{2} + 2 x} d x

g (t) = t^{2} - \frac{7}{t ^{3}}

x \to 0 lim (\frac{( 343 + x ) ^{\frac{1}{3}} - 7}{x})

\frac{\partial}{\partial x} (- 2 x^{2})

1 + sin (π \cdot t)

What is the integral of 1/(2x^2-5x+7) ?
The integral of 1/(2x^2-5x+7) is 2/(sqrt(31))arctan((4x-5)/(sqrt(31)))+C