What is the integral of (3x-1)/(4x^2-4x+17) ?

The integral of (3x-1)/(4x^2-4x+17) is 3/8 ln|4x^2-4x+17|+1/16 arctan(1/4 (2x-1))+C

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integral of (3x-1)/(4x^2-4x+17)

\int \frac{3 x - 1}{4 x ^{2} - 4 x + 17} d x

\frac{3}{8} ln 4 x^{2} - 4 x + 17 + \frac{1}{16} arctan (\frac{1}{4} (2 x - 1)) + C

Solution steps

\int \frac{3 x - 1}{4 x ^{2} - 4 x + 17} d x

Complete the square

4 x^{2} - 4 x + 17 : 4 (x - \frac{1}{2})^{2} + 16

= \int \frac{3 x - 1}{4 ( x - \frac{1}{2} ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{3 ( u + \frac{1}{2} ) - 1}{4 u ^{2} + 16} d u

= \int \frac{3 ( u + \frac{1}{2} ) - 1}{4 u ^{2} + 16} d u

Expand

\frac{3 ( u + \frac{1}{2} ) - 1}{4 u ^{2} + 16} : \frac{3 ( u + \frac{1}{2} )}{4 u ^{2} + 16} - \frac{1}{4 u ^{2} + 16}

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 + \frac{1}{2} )}{4 u ^{2} + 16} d u - \int \frac{1}{4 u ^{2} + 16} d u

\int \frac{3 ( u + \frac{1}{2} )}{4 u ^{2} + 16} d u = 3 (\frac{1}{8} ln 4 u^{2} + 16 + \frac{1}{16} arctan (\frac{u}{2}))

\int \frac{1}{4 u ^{2} + 16} d u = \frac{1}{8} arctan (\frac{u}{2})

= 3 (\frac{1}{8} ln 4 u^{2} + 16 + \frac{1}{16} arctan (\frac{u}{2})) - \frac{1}{8} arctan (\frac{u}{2})

Substitute back

u = x - \frac{1}{2}

= 3 (\frac{1}{8} ln 4 (x - \frac{1}{2})^{2} + 16 + \frac{1}{16} arctan (\frac{x - \frac{1}{2}}{2})) - \frac{1}{8} arctan (\frac{x - \frac{1}{2}}{2})

Simplify

3 (\frac{1}{8} ln 4 (x - \frac{1}{2})^{2} + 16 + \frac{1}{16} arctan (\frac{x - \frac{1}{2}}{2})) - \frac{1}{8} arctan (\frac{x - \frac{1}{2}}{2}) : \frac{3}{8} ln 4 x^{2} - 4 x + 17 + \frac{1}{16} arctan (\frac{1}{4} (2 x - 1))

= \frac{3}{8} ln 4 x^{2} - 4 x + 17 + \frac{1}{16} arctan (\frac{1}{4} (2 x - 1))

Add a constant to the solution

= \frac{3}{8} ln 4 x^{2} - 4 x + 17 + \frac{1}{16} arctan (\frac{1}{4} (2 x - 1)) + C

\frac{d y}{d x} = \frac{( 3 - y )}{2 x}

100 y^{^{''}} + 20 y^{^{'}} + y = 0

\int - x^{- 3} d x

x \to 1 + lim (x - 2)

\frac{\partial}{\partial x} (x y^{2} e^{x^{2}})

What is the integral of (3x-1)/(4x^2-4x+17) ?
The integral of (3x-1)/(4x^2-4x+17) is 3/8 ln|4x^2-4x+17|+1/16 arctan(1/4 (2x-1))+C