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

The integral of 1/(x^2-2x+50) is 1/7 arctan((x-1)/7)+C

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

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

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

Solution steps

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

Complete the square

x^{2} - 2 x + 50 : (x - 1)^{2} + 49

= \int \frac{1}{( x - 1 ) ^{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 - 1}{7})

Add a constant to the solution

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

\int \frac{1}{x - 2 x ^{\frac{2}{3}} - 3 x ^{\frac{1}{3}}} d x

\int x 3 x^{2} - 5 d x

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

\int \frac{1}{3 x ^{2} + 6 x - 4} d x

n = 1 \sum \infty \frac{1}{n + 5}

What is the integral of 1/(x^2-2x+50) ?
The integral of 1/(x^2-2x+50) is 1/7 arctan((x-1)/7)+C