What is the integral of 1/(x^2-10x+41) ?

The integral of 1/(x^2-10x+41) is 1/4 arctan((x-5)/4)+C

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integral of 1/(x^2-10x+41)

\int \frac{1}{x ^{2} - 10 x + 41} d x

\frac{1}{4} arctan (\frac{x - 5}{4}) + C

Solution steps

\int \frac{1}{x ^{2} - 10 x + 41} d x

Complete the square

x^{2} - 10 x + 41 : (x - 5)^{2} + 16

= \int \frac{1}{( x - 5 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 16} d u

= \int \frac{1}{u ^{2} + 16} d u

Apply Integral Substitution

= \int \frac{1}{4 ( 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}{4} \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}{4} arctan (v)

Substitute back

= \frac{1}{4} arctan (\frac{x - 5}{4})

Add a constant to the solution

= \frac{1}{4} arctan (\frac{x - 5}{4}) + C

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What is the integral of 1/(x^2-10x+41) ?
The integral of 1/(x^2-10x+41) is 1/4 arctan((x-5)/4)+C