What is the integral of 1/(x^2+6x+58) ?

The integral of 1/(x^2+6x+58) is 1/7 arctan((x+3)/7)+C

Popular >

integral of 1/(x^2+6x+58)

\int \frac{1}{x ^{2} + 6 x + 58} d x

\frac{1}{7} arctan (\frac{x + 3}{7}) + C

Solution steps

\int \frac{1}{x ^{2} + 6 x + 58} d x

Complete the square

x^{2} + 6 x + 58 : (x + 3)^{2} + 49

= \int \frac{1}{( x + 3 ) ^{2} + 49} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 49} d u

= \int \frac{1}{u ^{2} + 49} d u

Apply Integral Substitution

= \int \frac{1}{7 ( 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}{7} \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}{7} arctan (v)

Substitute back

= \frac{1}{7} arctan (\frac{x + 3}{7})

Add a constant to the solution

= \frac{1}{7} arctan (\frac{x + 3}{7}) + C

f (t) = sin^{2} (e^{s i n^{2} (t)})

\frac{2}{x}, at x = \frac{1}{4}

y = \frac{- 2 x}{x ^{2} + 1}, (1, - 1)

x \to 5 lim (3 x^{2} - 2 x - 7)

f (x) = 2 x^{2} + 5 x - 3, (1, 4)

What is the integral of 1/(x^2+6x+58) ?
The integral of 1/(x^2+6x+58) is 1/7 arctan((x+3)/7)+C