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

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

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

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

\frac{( x ^{2} - 1 ) ^{2}}{x} - 2 (- \frac{( x ^{2} - 1 ) ( x ^{2} + 1 )}{x} + \frac{4 x ^{3}}{3}) + C

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{2 ( x ^{2} - 1 ) ( x ^{2} + 1 )}{x ^{2}} d x = 2 (- \frac{( x ^{2} - 1 ) ( x ^{2} + 1 )}{x} + \frac{4 x ^{3}}{3})

= \frac{( x ^{2} - 1 ) ^{2}}{x} - 2 (- \frac{( x ^{2} - 1 ) ( x ^{2} + 1 )}{x} + \frac{4 x ^{3}}{3})

Add a constant to the solution

= \frac{( x ^{2} - 1 ) ^{2}}{x} - 2 (- \frac{( x ^{2} - 1 ) ( x ^{2} + 1 )}{x} + \frac{4 x ^{3}}{3}) + C

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