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

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

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

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

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

Solution steps

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

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

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

Apply u-substitution

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

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

- ln (cos (x))

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x \to 0 lim (\frac{arctan ( x ) - x}{8 x ^{3}})

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x \to - 2 lim (x^{2} - 3 x + 5)

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