integral of (4x-7)/(sqrt(x^2+6x+5))

Lösung

\int \frac{4 x - 7}{x ^{2} + 6 x + 5} d x

4 x^{2} + 6 x + 5 - 19 ln (\frac{1}{2} x^{2} + 6 x + 5 + x + 3) + C

Schritte zur Lösung

\int \frac{4 x - 7}{x ^{2} + 6 x + 5} d x

Vervollständige das Quadrat

x^{2} + 6 x + 5 : (x + 3)^{2} - 4

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

Wende U-Substitution an

= \int \frac{4 u - 19}{u ^{2} - 4} d u

Multipliziere aus

\frac{4 u - 19}{u ^{2} - 4} : \frac{4 u}{u ^{2} - 4} - \frac{19}{u ^{2} - 4}

= \int \frac{4 u}{u ^{2} - 4} - \frac{19}{u ^{2} - 4} d u

Wende die Summenregel an:

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

= \int \frac{4 u}{u ^{2} - 4} d u - \int \frac{19}{u ^{2} - 4} d u

\int \frac{4 u}{u ^{2} - 4} d u = 4 u^{2} - 4

\int \frac{19}{u ^{2} - 4} d u = 19 ln (\frac{1}{2} u^{2} - 4 + u)

= 4 u^{2} - 4 - 19 ln (\frac{1}{2} u^{2} - 4 + u)

Setze in

u = x + 3

ein

= 4 (x + 3)^{2} - 4 - 19 ln (\frac{1}{2} (x + 3)^{2} - 4 + x + 3)

Vereinfache

4 (x + 3)^{2} - 4 - 19 ln (\frac{1}{2} (x + 3)^{2} - 4 + x + 3) : 4 x^{2} + 6 x + 5 - 19 ln (\frac{1}{2} x^{2} + 6 x + 5 + x + 3)

= 4 x^{2} + 6 x + 5 - 19 ln (\frac{1}{2} x^{2} + 6 x + 5 + x + 3)

Füge eine Konstante zur Lösung hinzu

= 4 x^{2} + 6 x + 5 - 19 ln (\frac{1}{2} x^{2} + 6 x + 5 + x + 3) + C