What is the integral of ((10y+11)/(4y^2-4y+5)) ?

The integral of ((10y+11)/(4y^2-4y+5)) is 1/4 (5ln|y^2-y+5/4 |+16arctan(y-1/2))+C

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integral of ((10y+11)/(4y^2-4y+5))

\int (\frac{10 y + 11}{4 y ^{2} - 4 y + 5}) d y

\frac{1}{4} (5 ln y^{2} - y + \frac{5}{4} + 16 arctan (y - \frac{1}{2})) + C

Solution steps

\int \frac{10 y + 11}{4 y ^{2} - 4 y + 5} d y

Complete the square

4 y^{2} - 4 y + 5 : 4 (y - \frac{1}{2})^{2} + 4

= \int \frac{10 y + 11}{4 ( y - \frac{1}{2} ) ^{2} + 4} d y

Apply u-substitution

: \int \frac{5 u + 8}{2 ( u ^{2} + 1 )} d u

= \int \frac{5 u + 8}{2 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{5 u + 8}{u ^{2} + 1} d u

Expand

\frac{5 u + 8}{u ^{2} + 1} : \frac{5 u}{u ^{2} + 1} + \frac{8}{u ^{2} + 1}

Apply the Sum Rule:

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

= \frac{1}{2} (\int \frac{5 u}{u ^{2} + 1} d u + \int \frac{8}{u ^{2} + 1} d u)

\int \frac{5 u}{u ^{2} + 1} d u = \frac{5}{2} ln u^{2} + 1

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

= \frac{1}{2} (\frac{5}{2} ln u^{2} + 1 + 8 arctan (u))

Substitute back

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

= \frac{1}{2} (\frac{5}{2} ln (y - \frac{1}{2})^{2} + 1 + 8 arctan (y - \frac{1}{2}))

Simplify

\frac{1}{2} (\frac{5}{2} ln (y - \frac{1}{2})^{2} + 1 + 8 arctan (y - \frac{1}{2})) : \frac{1}{4} (5 ln y^{2} - y + \frac{5}{4} + 16 arctan (y - \frac{1}{2}))

= \frac{1}{4} (5 ln y^{2} - y + \frac{5}{4} + 16 arctan (y - \frac{1}{2}))

Add a constant to the solution

= \frac{1}{4} (5 ln y^{2} - y + \frac{5}{4} + 16 arctan (y - \frac{1}{2})) + C

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What is the integral of ((10y+11)/(4y^2-4y+5)) ?
The integral of ((10y+11)/(4y^2-4y+5)) is 1/4 (5ln|y^2-y+5/4 |+16arctan(y-1/2))+C