integral of 1/(4+4x^2)

Lösung

\int \frac{1}{4 + 4 x ^{2}} d x

\frac{1}{4} arctan (x) + C

Schritte zur Lösung

\int \frac{1}{4 + 4 x ^{2}} d x

\frac{1}{4 + 4 x ^{2}} = \frac{1}{4} \frac{1}{1 + x ^{2}}

= \int \frac{1}{4} \cdot \frac{1}{1 + x ^{2}} d x

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int \frac{1}{1 + x ^{2}} d x

Nutze das gemeinsame Integral :

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

= \frac{1}{4} arctan (x)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} arctan (x) + C

\int \frac{1}{( x + 4 ) ^{2} ( x - 3 )} d x

t an g e n t f (x) = x^{2} + 7, (1, 8)

\frac{d}{d x} (ln (arccos (x)))

2 y^{^{''}} + 5 y^{^{'}} - 18 y = 0

\frac{d x}{d t} = t^{\frac{3}{2}}, x (3) = 7