What is the integral of 3/(x^2-8x+25) ?

The integral of 3/(x^2-8x+25) is arctan(1/3 (x-4))+C

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integral of 3/(x^2-8x+25)

\int \frac{3}{x ^{2} - 8 x + 25} d x

arctan (\frac{1}{3} (x - 4)) + C

Solution steps

\int \frac{3}{x ^{2} - 8 x + 25} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{1}{x ^{2} - 8 x + 25} d x

Complete the square

x^{2} - 8 x + 25 : (x - 4)^{2} + 9

= 3 \cdot \int \frac{1}{( x - 4 ) ^{2} + 9} d x

Apply u-substitution

= 3 \cdot \int \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

= 3 \cdot \int \frac{1}{3 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 3 \cdot \frac{1}{3} arctan (v)

Substitute back

= 3 \cdot \frac{1}{3} arctan (\frac{x - 4}{3})

Simplify

3 \cdot \frac{1}{3} arctan (\frac{x - 4}{3}) : arctan (\frac{1}{3} (x - 4))

= arctan (\frac{1}{3} (x - 4))

Add a constant to the solution

= arctan (\frac{1}{3} (x - 4)) + C

What is the integral of 3/(x^2-8x+25) ?
The integral of 3/(x^2-8x+25) is arctan(1/3 (x-4))+C