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

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

Solution steps

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

(x + 1)^{2} + 4 = 4 (\frac{( x + 1 ) ^{2}}{4} + 1)

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

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{( x + 1 ) ^{2}}{4} = (\frac{1}{2} (x + 1))^{2}

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{4} \cdot 2 arctan (u)

Substitute back

u = \frac{1}{2} (x + 1)

= \frac{1}{4} \cdot 2 arctan (\frac{1}{2} (x + 1))

Simplify

\frac{1}{4} \cdot 2 arctan (\frac{1}{2} (x + 1)) : \frac{1}{2} arctan (\frac{1}{2} (x + 1))

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

Add a constant to the solution

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

\int x^{2} x^{3} + 33 d x

\frac{d y}{d x} + y^{5} x + 2 y = 0

\int \frac{1}{x ^{5} x ^{2} - 1} d x

x^{2} - 3 x

\frac{d}{d x} (- \frac{( a x + b - d ) ( x ^{2} - 1 )}{x ^{2}})

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