integral of ((2x-3))/(x^2+6x+13)

Lösung

\int \frac{( 2 x - 3 )}{x ^{2} + 6 x + 13} d x

ln \frac{1}{4} (x + 3)^{2} + 1 - \frac{9}{2} arctan (\frac{1}{2} (x + 3)) + C

Schritte zur Lösung

\int \frac{2 x - 3}{x ^{2} + 6 x + 13} d x

Vervollständige das Quadrat

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

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

Wende U-Substitution an

= \int \frac{2 u - 9}{u ^{2} + 4} d u

Wende integrale Substitution an

= \int \frac{4 v - 9}{2 ( v ^{2} + 1 )} d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int \frac{4 v - 9}{v ^{2} + 1} d v

Multipliziere aus

\frac{4 v - 9}{v ^{2} + 1} : \frac{4 v}{v ^{2} + 1} - \frac{9}{v ^{2} + 1}

Wende die Summenregel an:

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

= \frac{1}{2} (\int \frac{4 v}{v ^{2} + 1} d v - \int \frac{9}{v ^{2} + 1} d v)

\int \frac{4 v}{v ^{2} + 1} d v = 2 ln v^{2} + 1

\int \frac{9}{v ^{2} + 1} d v = 9 arctan (v)

= \frac{1}{2} (2 ln v^{2} + 1 - 9 arctan (v))

Ersetze zurück

= \frac{1}{2} (2 ln (\frac{x + 3}{2})^{2} + 1 - 9 arctan (\frac{x + 3}{2}))

Vereinfache

\frac{1}{2} (2 ln (\frac{x + 3}{2})^{2} + 1 - 9 arctan (\frac{x + 3}{2})) : ln \frac{1}{4} (x + 3)^{2} + 1 - \frac{9}{2} arctan (\frac{1}{2} (x + 3))

= ln \frac{1}{4} (x + 3)^{2} + 1 - \frac{9}{2} arctan (\frac{1}{2} (x + 3))

Füge eine Konstante zur Lösung hinzu

= ln \frac{1}{4} (x + 3)^{2} + 1 - \frac{9}{2} arctan (\frac{1}{2} (x + 3)) + C

\int \frac{x + 3}{( x ^{2} + 2 x + 5 ) ^{2}} d x

\int \frac{t ^{3}}{t ^{2} - 9} d t

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

\int 2 tan^{2} (2 x) d x

\int \frac{sin ( 15 x )}{1 + cos ^{2} ( 15 x )} d x