What is the integral of (9x^2+25)((7x-4)/(x^2)) ?

The integral of (9x^2+25)((7x-4)/(x^2)) is (63x^2)/2+175ln|x|-36x+(100)/x+C

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integral of (9x^2+25)((7x-4)/(x^2))

\int (9 x^{2} + 25) (\frac{7 x - 4}{x ^{2}}) d x

\frac{63 x ^{2}}{2} + 175 ln ∣ x ∣ - 36 x + \frac{100}{x} + C

Solution steps

\int (9 x^{2} + 25) \frac{7 x - 4}{x ^{2}} d x

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \int \frac{( 7 x - 4 ) ( 9 x ^{2} + 25 )}{x ^{2}} d x

Expand

\frac{( 7 x - 4 ) ( 9 x ^{2} + 25 )}{x ^{2}} : 63 x + \frac{175}{x} - 36 - \frac{100}{x ^{2}}

= \int 63 x + \frac{175}{x} - 36 - \frac{100}{x ^{2}} 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 63 x d x + \int \frac{175}{x} d x - \int 36 d x - \int \frac{100}{x ^{2}} d x

\int 63 x d x = \frac{63 x ^{2}}{2}

\int \frac{175}{x} d x = 175 ln ∣ x ∣

\int 36 d x = 36 x

\int \frac{100}{x ^{2}} d x = - \frac{100}{x}

= \frac{63 x ^{2}}{2} + 175 ln ∣ x ∣ - 36 x - (- \frac{100}{x})

Simplify

= \frac{63 x ^{2}}{2} + 175 ln ∣ x ∣ - 36 x + \frac{100}{x}

Add a constant to the solution

= \frac{63 x ^{2}}{2} + 175 ln ∣ x ∣ - 36 x + \frac{100}{x} + C

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What is the integral of (9x^2+25)((7x-4)/(x^2)) ?
The integral of (9x^2+25)((7x-4)/(x^2)) is (63x^2)/2+175ln|x|-36x+(100)/x+C