What is the integral of 1/(6x^2+36x+78) ?

The integral of 1/(6x^2+36x+78) is 1/12 arctan((x+3)/2)+C

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integral of 1/(6x^2+36x+78)

\int \frac{1}{6 x ^{2} + 36 x + 78} d x

\frac{1}{12} arctan (\frac{x + 3}{2}) + C

Solution steps

\int \frac{1}{6 x ^{2} + 36 x + 78} d x

Complete the square

6 x^{2} + 36 x + 78 : 6 (x + 3)^{2} + 24

= \int \frac{1}{6 ( x + 3 ) ^{2} + 24} d x

Apply u-substitution

: \int \frac{1}{6 u ^{2} + 24} d u

= \int \frac{1}{6 u ^{2} + 24} d u

Apply Integral Substitution

= \int \frac{1}{12 ( 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}{12} \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}{12} arctan (v)

Substitute back

= \frac{1}{12} arctan (\frac{6}{2 6} (x + 3))

Simplify

\frac{1}{12} arctan (\frac{6}{2 6} (x + 3)) : \frac{1}{12} arctan (\frac{x + 3}{2})

= \frac{1}{12} arctan (\frac{x + 3}{2})

Add a constant to the solution

= \frac{1}{12} arctan (\frac{x + 3}{2}) + C

tan (x) sin (y) d x + cos (x) cot (y) d y = 0

\int \frac{2 x ^{2} - x + 4}{x ( x ^{2} + 4 )} d x

y^{^{''}} - 2 y^{^{'}} + y = t e^{t}, y (0) = 1, y^{^{'}} (0) = 1

\int \frac{20}{x} d x

x \to 0 lim (5 + 2 e^{\frac{- 5}{x}})

What is the integral of 1/(6x^2+36x+78) ?
The integral of 1/(6x^2+36x+78) is 1/12 arctan((x+3)/2)+C