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

The integral of ((1-3x)^2)/(x^3) is -1/(2x^2)+6/x+9ln|x|+C

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

integral

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

- \frac{1}{2 x ^{2}} + \frac{6}{x} + 9 ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{( 1 - 3 x ) ^{2}}{x ^{3}} : \frac{1}{x ^{3}} - \frac{6}{x ^{2}} + \frac{9}{x}

= \int \frac{1}{x ^{3}} - \frac{6}{x ^{2}} + \frac{9}{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 ^{3}} d x - \int \frac{6}{x ^{2}} d x + \int \frac{9}{x} d x

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

\int \frac{6}{x ^{2}} d x = - \frac{6}{x}

\int \frac{9}{x} d x = 9 ln ∣ x ∣

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

Simplify

= - \frac{1}{2 x ^{2}} + \frac{6}{x} + 9 ln ∣ x ∣

Add a constant to the solution

= - \frac{1}{2 x ^{2}} + \frac{6}{x} + 9 ln ∣ x ∣ + C

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

sin^{2} (x) + cos^{2} (x)

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

sin^{2} (x) sin^{3} (x)

\frac{sin ^{2} ( x )}{2}

What is the integral of ((1-3x)^2)/(x^3) ?
The integral of ((1-3x)^2)/(x^3) is -1/(2x^2)+6/x+9ln|x|+C