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

The integral of 1/x ((2x^2-3x)^2+4) is x^4-4x^3+(9x^2)/2+4ln|x|+C

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

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

x^{4} - 4 x^{3} + \frac{9 x ^{2}}{2} + 4 ln ∣ x ∣ + C

Solution steps

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

Simplify

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

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

Expand

\frac{( 2 x ^{2} - 3 x ) ^{2} + 4}{x} : 4 x^{3} - 12 x^{2} + 9 x + \frac{4}{x}

= \int 4 x^{3} - 12 x^{2} + 9 x + \frac{4}{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 4 x^{3} d x - \int 12 x^{2} d x + \int 9 x d x + \int \frac{4}{x} d x

\int 4 x^{3} d x = x^{4}

\int 12 x^{2} d x = 4 x^{3}

\int 9 x d x = \frac{9 x ^{2}}{2}

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= x^{4} - 4 x^{3} + \frac{9 x ^{2}}{2} + 4 ln ∣ x ∣

Add a constant to the solution

= x^{4} - 4 x^{3} + \frac{9 x ^{2}}{2} + 4 ln ∣ x ∣ + C

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\int \frac{x ^{2} - 4 x + 4}{x ^{2} - 4 x + 29} d x

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\int \frac{x ^{3}}{( x ^{4} - 5 ) ^{4}} d x

What is the integral of 1/x ((2x^2-3x)^2+4) ?
The integral of 1/x ((2x^2-3x)^2+4) is x^4-4x^3+(9x^2)/2+4ln|x|+C