What is the integral of 1/(x^2+4x+68) ?

The integral of 1/(x^2+4x+68) is 1/8 arctan((x+2)/8)+C

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integral of 1/(x^2+4x+68)

\int \frac{1}{x ^{2} + 4 x + 68} d x

\frac{1}{8} arctan (\frac{x + 2}{8}) + C

Solution steps

\int \frac{1}{x ^{2} + 4 x + 68} d x

Complete the square

x^{2} + 4 x + 68 : (x + 2)^{2} + 64

= \int \frac{1}{( x + 2 ) ^{2} + 64} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 64} d u

= \int \frac{1}{u ^{2} + 64} d u

Apply Integral Substitution

= \int \frac{1}{8 ( 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}{8} \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}{8} arctan (v)

Substitute back

= \frac{1}{8} arctan (\frac{x + 2}{8})

Add a constant to the solution

= \frac{1}{8} arctan (\frac{x + 2}{8}) + C

\frac{\partial}{\partial x} (arctan (\frac{x}{y}))

\frac{d}{d x} (4 x^{3} - 12 x)

\int_{3}^{5} \frac{2 x ^{2} + x + 4}{x - 1} d x

\int_{1}^{e} \frac{( 1 + ln ( x ) ) ^{2}}{x} d x

\int \frac{tan ( ln ( x ) )}{x} d x

What is the integral of 1/(x^2+4x+68) ?
The integral of 1/(x^2+4x+68) is 1/8 arctan((x+2)/8)+C