What is the integral of ((7x-4)(3x+5)^2)/(x^2) ?

The integral of ((7x-4)(3x+5)^2)/(x^2) is (63x^2)/2+174x+55ln|x|+(100)/x+C

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integral of ((7x-4)(3x+5)^2)/(x^2)

\int \frac{( 7 x - 4 ) ( 3 x + 5 ) ^{2}}{x ^{2}} d x

\frac{63 x ^{2}}{2} + 174 x + 55 ln ∣ x ∣ + \frac{100}{x} + C

Solution steps

\int \frac{( 7 x - 4 ) ( 3 x + 5 ) ^{2}}{x ^{2}} d x

Expand

\frac{( 7 x - 4 ) ( 3 x + 5 ) ^{2}}{x ^{2}} : 63 x + 174 + \frac{55}{x} - \frac{100}{x ^{2}}

= \int 63 x + 174 + \frac{55}{x} - \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 174 d x + \int \frac{55}{x} d x - \int \frac{100}{x ^{2}} d x

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

\int 174 d x = 174 x

\int \frac{55}{x} d x = 55 ln ∣ x ∣

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

= \frac{63 x ^{2}}{2} + 174 x + 55 ln ∣ x ∣ - (- \frac{100}{x})

Simplify

= \frac{63 x ^{2}}{2} + 174 x + 55 ln ∣ x ∣ + \frac{100}{x}

Add a constant to the solution

= \frac{63 x ^{2}}{2} + 174 x + 55 ln ∣ x ∣ + \frac{100}{x} + C

\frac{\partial ^{2}}{\partial y ^{2}} (\frac{x}{x ^{2} + y ^{2}})

\int sin^{5} (2 x) cos^{2} (2 x) d x

f (x) = 2 x^{2} + 7

\frac{d y}{d x}, y = 3 x^{5}

\frac{\partial}{\partial y} (\frac{x}{x ^{4} - y ^{7}})

What is the integral of ((7x-4)(3x+5)^2)/(x^2) ?
The integral of ((7x-4)(3x+5)^2)/(x^2) is (63x^2)/2+174x+55ln|x|+(100)/x+C