What is the integral of (3x+1)/(x^2-10x+28) ?

The integral of (3x+1)/(x^2-10x+28) is 3/2 ln|x^2-10x+28|+(16)/(sqrt(3))arctan(1/(sqrt(3))(x-5))+C

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integral of (3x+1)/(x^2-10x+28)

\int \frac{3 x + 1}{x ^{2} - 10 x + 28} d x

\frac{3}{2} ln x^{2} - 10 x + 28 + \frac{16}{3} arctan (\frac{1}{3} (x - 5)) + C

Solution steps

\int \frac{3 x + 1}{x ^{2} - 10 x + 28} d x

Complete the square

x^{2} - 10 x + 28 : (x - 5)^{2} + 3

= \int \frac{3 x + 1}{( x - 5 ) ^{2} + 3} d x

Apply u-substitution

: \int \frac{3 u + 16}{u ^{2} + 3} d u

= \int \frac{3 u + 16}{u ^{2} + 3} d u

Expand

\frac{3 u + 16}{u ^{2} + 3} : \frac{3 u}{u ^{2} + 3} + \frac{16}{u ^{2} + 3}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3 u}{u ^{2} + 3} d u + \int \frac{16}{u ^{2} + 3} d u

\int \frac{3 u}{u ^{2} + 3} d u = \frac{3}{2} ln u^{2} + 3

\int \frac{16}{u ^{2} + 3} d u = \frac{16}{3} arctan (\frac{u}{3})

= \frac{3}{2} ln u^{2} + 3 + \frac{16}{3} arctan (\frac{u}{3})

Substitute back

u = x - 5

= \frac{3}{2} ln (x - 5)^{2} + 3 + \frac{16}{3} arctan (\frac{x - 5}{3})

Simplify

\frac{3}{2} ln (x - 5)^{2} + 3 + \frac{16}{3} arctan (\frac{x - 5}{3}) : \frac{3}{2} ln x^{2} - 10 x + 28 + \frac{16}{3} arctan (\frac{1}{3} (x - 5))

= \frac{3}{2} ln x^{2} - 10 x + 28 + \frac{16}{3} arctan (\frac{1}{3} (x - 5))

Add a constant to the solution

= \frac{3}{2} ln x^{2} - 10 x + 28 + \frac{16}{3} arctan (\frac{1}{3} (x - 5)) + C

200 (e)^{0.03 x}

y = x^{4} - 3 x^{2} + 3, y = x^{2}

y = 5 x^{2} - x^{3}, (1, 4)

x \to 2 lim (\frac{2}{x})

What is the integral of (3x+1)/(x^2-10x+28) ?
The integral of (3x+1)/(x^2-10x+28) is 3/2 ln|x^2-10x+28|+(16)/(sqrt(3))arctan(1/(sqrt(3))(x-5))+C