What is the integral of 1/((x-3)^2+3) ?

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

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integral of 1/((x-3)^2+3)

\int \frac{1}{( x - 3 ) ^{2} + 3} d x

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

Solution steps

\int \frac{1}{( x - 3 ) ^{2} + 3} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 3} d u

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

Apply Integral Substitution

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

Use the common integral:

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

= \frac{1}{3} arctan (v)

Substitute back

= \frac{1}{3} arctan (\frac{x - 3}{3})

Add a constant to the solution

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

x \to - 10 lim (7)

\int (y cos (x y)) d x

x \to - 7 + lim (\frac{x + 8}{x + 7})

\frac{d}{d x} (5 ln (4 x^{2}))

x^{8} + y^{7} = 9

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