integral from 1 to 3 of 1/((x+1)^2)

Lösung

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

\frac{1}{4}

Dezimale

0.25

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{2}^{4} \frac{1}{u ^{2}} d u

Wende die Potenzregel an

= [- \frac{1}{u}]_{2}^{4}

Berechne die Grenzen:

\frac{1}{4}

= \frac{1}{4}

\int_{\frac{π}{3}}^{\frac{2 π}{3}} sin (x) d x

\int_{0}^{4} x sin (3 x) d x

\int_{0}^{2 x} 2 x y^{3} - x d y

\int_{- 1}^{1} ∣ x ∣ sin (x) d x

\int_{0}^{\frac{π}{2}} x sin (3 x) d x