What is the integral of (x^{-1}+3)^3 ?

The integral of (x^{-1}+3)^3 is ((1+3x)^3}{x^2}-3/(2x^2)-\frac{18)/x+27ln|x|+C

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integral of (x^{-1}+3)^3

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

\frac{( 1 + 3 x ) ^{3}}{x ^{2}} - \frac{3}{2 x ^{2}} - \frac{18}{x} + 27 ln ∣ x ∣ + C

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Apply u-substitution

: \int - \frac{( u + 3 ) ^{3}}{u ^{2}} d u

= \int - \frac{( u + 3 ) ^{3}}{u ^{2}} d u

Take the constant out:

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

= - \int \frac{( u + 3 ) ^{3}}{u ^{2}} d u

Apply Integration By Parts

: - \frac{( u + 3 ) ^{3}}{u} - \int - \frac{3 ( u + 3 ) ^{2}}{u} d u

= - (- \frac{( u + 3 ) ^{3}}{u} - \int - \frac{3 ( u + 3 ) ^{2}}{u} d u)

\int - \frac{3 ( u + 3 ) ^{2}}{u} d u = - 3 (\frac{u ^{2}}{2} + 6 u + 9 ln ∣ u ∣)

= - (- \frac{( u + 3 ) ^{3}}{u} - (- 3 (\frac{u ^{2}}{2} + 6 u + 9 ln ∣ u ∣)))

Substitute back

u = \frac{1}{x}

= - (- \frac{( \frac{1}{x} + 3 ) ^{3}}{\frac{1}{x}} - (- 3 (\frac{( \frac{1}{x} ) ^{2}}{2} + 6 \cdot \frac{1}{x} + 9 ln \frac{1}{x})))

Simplify

- (- \frac{( \frac{1}{x} + 3 ) ^{3}}{\frac{1}{x}} - (- 3 (\frac{( \frac{1}{x} ) ^{2}}{2} + 6 \cdot \frac{1}{x} + 9 ln \frac{1}{x}))) : \frac{( 1 + 3 x ) ^{3}}{x ^{2}} - \frac{3}{2 x ^{2}} - \frac{18}{x} + 27 ln ∣ x ∣

= \frac{( 1 + 3 x ) ^{3}}{x ^{2}} - \frac{3}{2 x ^{2}} - \frac{18}{x} + 27 ln ∣ x ∣

Add a constant to the solution

= \frac{( 1 + 3 x ) ^{3}}{x ^{2}} - \frac{3}{2 x ^{2}} - \frac{18}{x} + 27 ln ∣ x ∣ + C

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What is the integral of (x^{-1}+3)^3 ?
The integral of (x^{-1}+3)^3 is ((1+3x)^3}{x^2}-3/(2x^2)-\frac{18)/x+27ln|x|+C