What is the integral of x/((x-3)^2+4^2) ?

The integral of x/((x-3)^2+4^2) is 1/2 ln|x^2-6x+25|+3/4 arctan(1/4 (x-3))+C

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integral of x/((x-3)^2+4^2)

\int \frac{x}{( x - 3 ) ^{2} + 4 ^{2}} d x

\frac{1}{2} ln x^{2} - 6 x + 25 + \frac{3}{4} arctan (\frac{1}{4} (x - 3)) + C

Solution steps

\int \frac{x}{( x - 3 ) ^{2} + 4 ^{2}} d x

Simplify

\frac{x}{( x - 3 ) ^{2} + 4 ^{2}} : \frac{x}{x ^{2} - 6 x + 25}

= \int \frac{x}{x ^{2} - 6 x + 25} d x

Complete the square

x^{2} - 6 x + 25 : (x - 3)^{2} + 16

= \int \frac{x}{( x - 3 ) ^{2} + 16} d x

Apply u-substitution

= \int \frac{u + 3}{u ^{2} + 16} d u

Expand

\frac{u + 3}{u ^{2} + 16} : \frac{u}{u ^{2} + 16} + \frac{3}{u ^{2} + 16}

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{u}{u ^{2} + 16} d u + \int \frac{3}{u ^{2} + 16} d u

\int \frac{u}{u ^{2} + 16} d u = \frac{1}{2} ln u^{2} + 16

\int \frac{3}{u ^{2} + 16} d u = \frac{3}{4} arctan (\frac{u}{4})

= \frac{1}{2} ln u^{2} + 16 + \frac{3}{4} arctan (\frac{u}{4})

Substitute back

u = x - 3

= \frac{1}{2} ln (x - 3)^{2} + 16 + \frac{3}{4} arctan (\frac{x - 3}{4})

Simplify

\frac{1}{2} ln (x - 3)^{2} + 16 + \frac{3}{4} arctan (\frac{x - 3}{4}) : \frac{1}{2} ln x^{2} - 6 x + 25 + \frac{3}{4} arctan (\frac{1}{4} (x - 3))

= \frac{1}{2} ln x^{2} - 6 x + 25 + \frac{3}{4} arctan (\frac{1}{4} (x - 3))

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 6 x + 25 + \frac{3}{4} arctan (\frac{1}{4} (x - 3)) + C

What is the integral of x/((x-3)^2+4^2) ?
The integral of x/((x-3)^2+4^2) is 1/2 ln|x^2-6x+25|+3/4 arctan(1/4 (x-3))+C