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

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

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

integral

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

- \frac{1}{x} + 3 ln ∣ x ∣ + 3 x + \frac{x ^{2}}{2} + C

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int 3 d x = 3 x

\int x d x = \frac{x ^{2}}{2}

= - \frac{1}{x} + 3 ln ∣ x ∣ + 3 x + \frac{x ^{2}}{2}

Add a constant to the solution

= - \frac{1}{x} + 3 ln ∣ x ∣ + 3 x + \frac{x ^{2}}{2} + C

x \cdot 3^{x}

3 x^{2} + 5

1 + cos^{2} (x)

x^{\frac{1}{3}}

\frac{1}{4 + x ^{2}}

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