What is area y= x/(x^2-4x+5),y= 3/(x^2-4x+5),x=0 ?

The answer to area y= x/(x^2-4x+5),y= 3/(x^2-4x+5),x=0 is 1/2 ln(5)-1/2 ln(2)+pi/4+arctan(2)

Popular >

area y= x/(x^2-4x+5),y= 3/(x^2-4x+5),x=0

area

y = \frac{x}{x ^{2} - 4 x + 5}, y = \frac{3}{x ^{2} - 4 x + 5}, x = 0

\frac{1}{2} ln (5) - \frac{1}{2} ln (2) + \frac{π}{4} + arctan (2)

Decimal Notation

2.35069 \dots

Solution steps

Apply the area formula:

\int_{0}^{3} \frac{x}{x ^{2} - 4 x + 5} - \frac{3}{x ^{2} - 4 x + 5} d x

Solve

\int_{0}^{3} \frac{x}{x ^{2} - 4 x + 5} - \frac{3}{x ^{2} - 4 x + 5} d x : \frac{1}{2} ln (5) - \frac{1}{2} ln (2) + \frac{π}{4} + arctan (2)

The area is:

= \frac{1}{2} ln (5) - \frac{1}{2} ln (2) + \frac{π}{4} + arctan (2)

y = \frac{x ^{3} + 1}{6 x}, x = 1

y^{^{'}} + 4 y = e^{- x}, y (0) = 10

\int_{0}^{6} \frac{1}{∣ x - 3 ∣} d x

\frac{d}{d x} (x (x^{2} + 3)^{3})

y^{^{''}} + 9 y = 2 x^{2} - 5

What is area y= x/(x^2-4x+5),y= 3/(x^2-4x+5),x=0 ?
The answer to area y= x/(x^2-4x+5),y= 3/(x^2-4x+5),x=0 is 1/2 ln(5)-1/2 ln(2)+pi/4+arctan(2)