integral of (x+6)/(sqrt(25-(x-5)^2))

Lösung

\int \frac{x + 6}{25 - ( x - 5 ) ^{2}} d x

- - x^{2} + 10 x + 11 arcsin (\frac{1}{5} x - 1) + C

Schritte zur Lösung

\int \frac{x + 6}{25 - ( x - 5 ) ^{2}} d x

Wende U-Substitution an

= \int \frac{u + 11}{25 - u ^{2}} d u

Trigonometrische Substitution anwenden

= \int 5 sin (v) + 11 d v

Wende die Summenregel an:

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

= \int 5 sin (v) d v + \int 11 d v

\int 5 sin (v) d v = - 5 cos (v)

\int 11 d v = 11 v

= - 5 cos (v) + 11 v

Ersetze zurück

= - 5 cos (arcsin (\frac{1}{5} (x - 5))) + 11 arcsin (\frac{1}{5} (x - 5))

Vereinfache

- 5 cos (arcsin (\frac{1}{5} (x - 5))) + 11 arcsin (\frac{1}{5} (x - 5)) : - - x^{2} + 10 x + 11 arcsin (\frac{1}{5} x - 1)

= - - x^{2} + 10 x + 11 arcsin (\frac{1}{5} x - 1)

Füge eine Konstante zur Lösung hinzu

= - - x^{2} + 10 x + 11 arcsin (\frac{1}{5} x - 1) + C

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

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

\int (x + \frac{sin ( x )}{x}) d x

in t e g r a l cos (π x)

\int 5 x^{2} cos (7 x) d x