What is the integral of 1/(x^2+10x+50) ?

The integral of 1/(x^2+10x+50) is 1/5 arctan((x+5)/5)+C

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integral of 1/(x^2+10x+50)

\int \frac{1}{x ^{2} + 10 x + 50} d x

\frac{1}{5} arctan (\frac{x + 5}{5}) + C

Solution steps

\int \frac{1}{x ^{2} + 10 x + 50} d x

Complete the square

x^{2} + 10 x + 50 : (x + 5)^{2} + 25

= \int \frac{1}{( x + 5 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 25} d u

= \int \frac{1}{u ^{2} + 25} d u

Apply Integral Substitution

= \int \frac{1}{5 ( 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}{5} \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}{5} arctan (v)

Substitute back

= \frac{1}{5} arctan (\frac{x + 5}{5})

Add a constant to the solution

= \frac{1}{5} arctan (\frac{x + 5}{5}) + C

f (x) = \frac{1 - 5 x}{1 + 5 x}

1 - e^{- x}

\frac{1}{e ^{x}}

\frac{d}{d x} (5 x^{5})

x \to 0 lim (cot (x) - csc (x))

What is the integral of 1/(x^2+10x+50) ?
The integral of 1/(x^2+10x+50) is 1/5 arctan((x+5)/5)+C