What is the integral of 1/((x-1)^2+4) ?

The integral of 1/((x-1)^2+4) is 1/2 arctan((x-1)/2)+C

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integral of 1/((x-1)^2+4)

\int \frac{1}{( x - 1 ) ^{2} + 4} d x

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

Solution steps

\int \frac{1}{( x - 1 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 4} d u

= \int \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

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

Substitute back

= \frac{1}{2} arctan (\frac{x - 1}{2})

Add a constant to the solution

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

\int_{0}^{π} t \cdot sin (3 t) d t

\frac{d}{d x} (e^{x} sin (y))

x \to - 1 lim (x^{2} - 1)

n = 0 \sum \infty \frac{2}{5 ^{n - 1}}

y = sin (sin (x)), (3 π, 0)

What is the integral of 1/((x-1)^2+4) ?
The integral of 1/((x-1)^2+4) is 1/2 arctan((x-1)/2)+C