integral from 1 to 5 of sqrt(x^2+4)

Lösung

\int_{1}^{5} x^{2} + 4 d x

4 (\frac{5 29 - 5}{8} + \frac{1}{2} ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4}))

Dezimale

14.67691 \dots

Schritte zur Lösung

\int_{1}^{5} x^{2} + 4 d x

Trigonometrische Substitution anwenden

= \int_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} 4 sec^{3} (u) d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} sec^{3} (u) d u

Wende integrale Reduktion an

= 4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} \cdot \int_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} sec (u) d u

\int_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} sec (u) d u = ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4})

= 4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4})

Vereinfache

4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4}) : 4 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4}))

= 4 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4}))

Berechne die Grenzen:

\frac{5 29 - 5}{8}

= 4 (\frac{5 29 - 5}{8} + \frac{1}{2} ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4}))

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