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

The integral of 8/(x^2-2x+2) is 8arctan(x-1)+C

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integral of 8/(x^2-2x+2)

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

8 arctan (x - 1) + C

Solution steps

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

Take the constant out:

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

= 8 \cdot \int \frac{1}{x ^{2} - 2 x + 2} d x

Complete the square

x^{2} - 2 x + 2 : (x - 1)^{2} + 1

= 8 \cdot \int \frac{1}{( x - 1 ) ^{2} + 1} d x

Apply u-substitution

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

Use the common integral:

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

= 8 arctan (u)

Substitute back

u = x - 1

= 8 arctan (x - 1)

Add a constant to the solution

= 8 arctan (x - 1) + C

\frac{\partial}{\partial x} (x y - x - 3 y)

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

\int \frac{7}{x ^{3}} d x

y = x^{10} e^{x}

y^{^{''}} - 5 y^{^{'}} + 6 y = t^{2}

What is the integral of 8/(x^2-2x+2) ?
The integral of 8/(x^2-2x+2) is 8arctan(x-1)+C