What is the integral of 1/(x^2+6x+16) ?

The integral of 1/(x^2+6x+16) is 1/(sqrt(7))arctan((x+3)/(sqrt(7)))+C

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integral of 1/(x^2+6x+16)

\int \frac{1}{x ^{2} + 6 x + 16} d x

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

Solution steps

\int \frac{1}{x ^{2} + 6 x + 16} d x

Complete the square

x^{2} + 6 x + 16 : (x + 3)^{2} + 7

= \int \frac{1}{( x + 3 ) ^{2} + 7} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 7} d u

= \int \frac{1}{u ^{2} + 7} 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

x \to 0 lim (\frac{sin ( x )}{9 x})

6 x^{2} y^{^{''}} + 13 x y^{^{'}} + y = 70 x + 2

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

\frac{d}{d x} (2 x (1 - x^{2})^{\frac{1}{2}})

x \frac{d y}{d x} + 2 y = 4 e^{x}

What is the integral of 1/(x^2+6x+16) ?
The integral of 1/(x^2+6x+16) is 1/(sqrt(7))arctan((x+3)/(sqrt(7)))+C