What is the integral of (1+1/x)^3 1/(x^2) ?

The integral of (1+1/x)^3 1/(x^2) is -((x+1)^4)/(4x^4)+C

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integral of (1+1/x)^3 1/(x^2)

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

- \frac{( x + 1 ) ^{4}}{4 x ^{4}} + C

Solution steps

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

Apply u-substitution

: \int - u^{3} d u

= \int - u^{3} d u

Take the constant out:

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

= - \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= - \frac{u ^{4}}{4}

Substitute back

u = 1 + \frac{1}{x}

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

Simplify

- \frac{( 1 + \frac{1}{x} ) ^{4}}{4} : - \frac{( x + 1 ) ^{4}}{4 x ^{4}}

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

Add a constant to the solution

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

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What is the integral of (1+1/x)^3 1/(x^2) ?
The integral of (1+1/x)^3 1/(x^2) is -((x+1)^4)/(4x^4)+C