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

Lösung

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

3 x^{2} - 2 x - 1

Schritte zur Lösung

\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

Wende die Summenregel an:

\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)

Vereinfache

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

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

\int_{0}^{\frac{π}{2}} (8 + 4 cos (x)) d x

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