What is the integral of (4x^3-8x^2+4x+3)/(4x^2-8x+5) ?

The integral of (4x^3-8x^2+4x+3)/(4x^2-8x+5) is (x^2)/2-1/8 ln|4x^2-8x+5|+arctan(2(x-1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x^3-8x^2+4x+3)/(4x^2-8x+5)

\int \frac{4 x ^{3} - 8 x ^{2} + 4 x + 3}{4 x ^{2} - 8 x + 5} d x

\frac{x ^{2}}{2} - \frac{1}{8} ln 4 x^{2} - 8 x + 5 + arctan (2 (x - 1)) + C

Solution steps

\int \frac{4 x ^{3} - 8 x ^{2} + 4 x + 3}{4 x ^{2} - 8 x + 5} d x

Long division

\frac{4 x ^{3} - 8 x ^{2} + 4 x + 3}{4 x ^{2} - 8 x + 5} : x + \frac{- x + 3}{4 x ^{2} - 8 x + 5}

= \int x + \frac{- x + 3}{4 x ^{2} - 8 x + 5} d x

Apply the Sum Rule:

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

= \int x d x + \int \frac{- x + 3}{4 x ^{2} - 8 x + 5} d x

\int x d x = \frac{x ^{2}}{2}

\int \frac{- x + 3}{4 x ^{2} - 8 x + 5} d x = - \frac{1}{8} ln 4 x^{2} - 8 x + 5 + arctan (2 (x - 1))

= \frac{x ^{2}}{2} - \frac{1}{8} ln 4 x^{2} - 8 x + 5 + arctan (2 (x - 1))

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{1}{8} ln 4 x^{2} - 8 x + 5 + arctan (2 (x - 1)) + C

\int (5 x^{4} - 3 x^{4} + 4 x^{2} - 2 + 6) d x

\int x 17 x - 14 d x

\int e^{x^{13}} d x

\int \frac{cos ( 9 x )}{sec ( 9 x ) - tan ( 9 x )} d x

\int \frac{tan ( y )}{sin ^{2} ( y )} d y

What is the integral of (4x^3-8x^2+4x+3)/(4x^2-8x+5) ?
The integral of (4x^3-8x^2+4x+3)/(4x^2-8x+5) is (x^2)/2-1/8 ln|4x^2-8x+5|+arctan(2(x-1))+C