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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

\frac{d}{d x} (x^{3} \cdot sin (x))

\int 9 x^{- 1} d x

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

f (x) = \frac{3}{( x + 5 ) ^{2}}, at x = 1

x \to 3 lim (\frac{∣ x + 7 ∣}{x - 3})

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