What is the integral of y=(x^2-cos(x)+1)/x ?

The integral of y=(x^2-cos(x)+1)/x is (x^2)/2-\Ci(x)+ln|x|+C

integral of y=(x^2-cos(x)+1)/x

integral

y = \frac{x ^{2} - cos ( x ) + 1}{x}

\frac{x ^{2}}{2} - Ci (x) + ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{x ^{2} - cos ( x ) + 1}{x} : x - \frac{cos ( x )}{x} + \frac{1}{x}

= \int x - \frac{cos ( x )}{x} + \frac{1}{x} 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{cos ( x )}{x} d x + \int \frac{1}{x} d x

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

\int \frac{cos ( x )}{x} d x = Ci (x)

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{2}}{2} - Ci (x) + ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - Ci (x) + ln ∣ x ∣ + C

in t e g r a l \frac{0.001}{( 0.001 x + 1 ) ^{2}}

in t e g r a l x - 3

in t e g r a l \frac{( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )}

in t e g r a l 1. 3^{x}

in t e g r a l \frac{3 x}{( x ^{2} + 4 x + 6 )}

What is the integral of y=(x^2-cos(x)+1)/x ?
The integral of y=(x^2-cos(x)+1)/x is (x^2)/2-\Ci(x)+ln|x|+C