What is the integral of 1/(x^2-14x+68) ?

The integral of 1/(x^2-14x+68) is 1/(sqrt(19))arctan((x-7)/(sqrt(19)))+C

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integral of 1/(x^2-14x+68)

\int \frac{1}{x ^{2} - 14 x + 68} d x

\frac{1}{19} arctan (\frac{x - 7}{19}) + C

Solution steps

\int \frac{1}{x ^{2} - 14 x + 68} d x

Complete the square

x^{2} - 14 x + 68 : (x - 7)^{2} + 19

= \int \frac{1}{( x - 7 ) ^{2} + 19} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 19} d u

= \int \frac{1}{u ^{2} + 19} d u

Apply Integral Substitution

= \int \frac{1}{19 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{19} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{19} arctan (v)

Substitute back

= \frac{1}{19} arctan (\frac{x - 7}{19})

Add a constant to the solution

= \frac{1}{19} arctan (\frac{x - 7}{19}) + C

\frac{d y}{d x} + (\frac{y}{x}) = 7 x^{3} y^{2}

\frac{\partial}{\partial x} (25 x^{\frac{3}{4}} y^{\frac{1}{4}})

\int - csc^{2} (x) d x

\frac{d y}{d t} = y

y = 2 x - 1

What is the integral of 1/(x^2-14x+68) ?
The integral of 1/(x^2-14x+68) is 1/(sqrt(19))arctan((x-7)/(sqrt(19)))+C