What is the integral of 1/((x+1)^2+9) ?

The integral of 1/((x+1)^2+9) is 1/3 arctan(1/3 (x+1))+C

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integral of 1/((x+1)^2+9)

\int \frac{1}{( x + 1 ) ^{2} + 9} d x

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

Solution steps

\int \frac{1}{( x + 1 ) ^{2} + 9} d x

(x + 1)^{2} + 9 = 9 (\frac{( x + 1 ) ^{2}}{9} + 1)

= \int \frac{1}{9 ( \frac{( x + 1 ) ^{2}}{9} + 1 )} d x

Apply exponent rule:

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

\frac{( x + 1 ) ^{2}}{9} = (\frac{1}{3} (x + 1))^{2}

= \int \frac{1}{9 ( ( \frac{1}{3} ( 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}{9} \cdot \int \frac{1}{( \frac{1}{3} ( x + 1 ) ) ^{2} + 1} d x

Apply u-substitution

: \int \frac{9}{u ^{2} + 9} d u

= \frac{1}{9} \cdot \int \frac{9}{u ^{2} + 9} d u

Take the constant out:

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

= \frac{1}{9} \cdot 9 \cdot \int \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

= \frac{1}{9} \cdot 9 \cdot \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}{9} \cdot 9 \cdot \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}{9} \cdot 9 \cdot \frac{1}{3} arctan (v)

Substitute back

= \frac{1}{9} \cdot 9 \cdot \frac{1}{3} arctan (\frac{x + 1}{3})

Simplify

\frac{1}{9} \cdot 9 \cdot \frac{1}{3} arctan (\frac{x + 1}{3}) : \frac{1}{3} arctan (\frac{1}{3} (x + 1))

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

Add a constant to the solution

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

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What is the integral of 1/((x+1)^2+9) ?
The integral of 1/((x+1)^2+9) is 1/3 arctan(1/3 (x+1))+C