What is the integral from 1 to x of (2x^2y^{-2}+2y) ?

The integral from 1 to x of (2x^2y^{-2}+2y) is 3x^2-2x-1

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integral from 1 to x of (2x^2y^{-2}+2y)

\int_{1}^{x} (2 x^{2} y^{- 2} + 2 y) d y

3 x^{2} - 2 x - 1

Solution steps

\int_{1}^{x} 2 x^{2} y^{- 2} + 2 y d y

2 x^{2} y^{- 2} = \frac{2 x ^{2}}{y ^{2}}

= \int_{1}^{x} 2 y + \frac{2 x ^{2}}{y ^{2}} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{x} 2 y d y + \int_{1}^{x} \frac{2 x ^{2}}{y ^{2}} d y

\int_{1}^{x} 2 y d y = x^{2} - 1

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

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

Simplify

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

= 3 x^{2} - 2 x - 1

\int_{2}^{e + 1} \frac{x}{x - 1} d x

\int_{0}^{1} 2 e^{\frac{x}{3} + 4} + 5 d x

\int_{0}^{x} x^{2} d x

\int_{0}^{\frac{π}{4}} \frac{1}{cos ^{2} ( x )} d x

\int_{1}^{\infty} x^{2} e^{- x} d x

What is the integral from 1 to x of (2x^2y^{-2}+2y) ?
The integral from 1 to x of (2x^2y^{-2}+2y) is 3x^2-2x-1