What is the integral of x/(x^2-4x+7) ?

The integral of x/(x^2-4x+7) is 1/2 ln|((x-2)^2)/3+1|+2/(sqrt(3))arctan(1/(sqrt(3))(x-2))+C

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integral of x/(x^2-4x+7)

\int \frac{x}{x ^{2} - 4 x + 7} d x

\frac{1}{2} ln \frac{( x - 2 ) ^{2}}{3} + 1 + \frac{2}{3} arctan (\frac{1}{3} (x - 2)) + C

Solution steps

\int \frac{x}{x ^{2} - 4 x + 7} d x

Complete the square

x^{2} - 4 x + 7 : (x - 2)^{2} + 3

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

Apply u-substitution

: \int \frac{u + 2}{u ^{2} + 3} d u

= \int \frac{u + 2}{u ^{2} + 3} d u

Apply Integral Substitution

= \int \frac{3 v + 2}{3 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{3 v + 2}{v ^{2} + 1} d v

Expand

\frac{3 v + 2}{v ^{2} + 1} : \frac{3 v}{v ^{2} + 1} + \frac{2}{v ^{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

= \frac{1}{3} (\int \frac{3 v}{v ^{2} + 1} d v + \int \frac{2}{v ^{2} + 1} d v)

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

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

= \frac{1}{3} (\frac{3}{2} ln v^{2} + 1 + 2 arctan (v))

Substitute back

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

Simplify

\frac{1}{3} (\frac{3}{2} ln (\frac{x - 2}{3})^{2} + 1 + 2 arctan (\frac{x - 2}{3})) : \frac{1}{2} ln \frac{( x - 2 ) ^{2}}{3} + 1 + \frac{2}{3} arctan (\frac{1}{3} (x - 2))

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

Add a constant to the solution

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

x \to 1 lim (ab s v a l)

\frac{d y}{d x} = \frac{( x ^{2} - y ^{2} )}{- x y}

\frac{4}{s ^{2} + 4}

n = 0 \sum \infty 3^{n} x^{n}

\int_{- 2}^{3} (8 x^{3} + 3 x - 1) d x

What is the integral of x/(x^2-4x+7) ?
The integral of x/(x^2-4x+7) is 1/2 ln|((x-2)^2)/3+1|+2/(sqrt(3))arctan(1/(sqrt(3))(x-2))+C