What is the integral of (x^2-6x+1)/(x^2+1) ?

The integral of (x^2-6x+1)/(x^2+1) is x-3ln|x^2+1|+C

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integral of (x^2-6x+1)/(x^2+1)

\int \frac{x ^{2} - 6 x + 1}{x ^{2} + 1} d x

x - 3 ln x^{2} + 1 + C

Solution steps

\int \frac{x ^{2} - 6 x + 1}{x ^{2} + 1} d x

Expand

\frac{x ^{2} - 6 x + 1}{x ^{2} + 1} : \frac{x ^{2}}{x ^{2} + 1} - \frac{6 x}{x ^{2} + 1} + \frac{1}{x ^{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

= \int \frac{x ^{2}}{x ^{2} + 1} d x - \int \frac{6 x}{x ^{2} + 1} d x + \int \frac{1}{x ^{2} + 1} d x

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

\int \frac{6 x}{x ^{2} + 1} d x = 3 ln x^{2} + 1

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

= - arctan (x) + x - 3 ln x^{2} + 1 + arctan (x)

Simplify

= x - 3 ln x^{2} + 1

Add a constant to the solution

= x - 3 ln x^{2} + 1 + C

\frac{d}{d x} ((4 - x^{2})^{\frac{1}{2}})

\int_{0}^{\infty} 6 e^{- x} d x

n = 0 \sum \infty (x - 9)^{n}

\frac{d}{d x} (2 + ln (x))

y^{^{'}} + 4 x y = x

What is the integral of (x^2-6x+1)/(x^2+1) ?
The integral of (x^2-6x+1)/(x^2+1) is x-3ln|x^2+1|+C