Integrale mit Substitution und partieller Integration

Aufgabenstellung

Berechnen Sie die folgenden Integrale. Geben Sie jeweils die verwendete Methode an.

(a) $\displaystyle\int 2x \cdot e^{x^2}\,dx$

Substitution: $u = x^2$ , $du = 2x\,dx$ :

$\int e^u\,du = e^u + C = e^{x^2} + C$

$\boxed{\int 2x \cdot e^{x^2}\,dx = e^{x^2} + C}$

(b) $\displaystyle\int \frac{6x}{(3x^2+1)^2}\,dx$

Substitution: $u = 3x^2 + 1$ , $du = 6x\,dx$ :

$\int \frac{1}{u^2}\,du = \int u^{-2}\,du = -u^{-1} + C = -\frac{1}{3x^2+1} + C$

$\boxed{\int \frac{6x}{(3x^2+1)^2}\,dx = -\frac{1}{3x^2+1} + C}$

(c) $\displaystyle\int \cos(x) \cdot e^{\sin(x)}\,dx$

Substitution: $u = \sin(x)$ , $du = \cos(x)\,dx$ :

$\int e^u\,du = e^u + C = e^{\sin(x)} + C$

$\boxed{\int \cos(x) \cdot e^{\sin(x)}\,dx = e^{\sin(x)} + C}$

$\displaystyle\int x \cdot e^x\,dx$

Wahl: $u = x$ , $v' = e^x$ → $u' = 1$ , $v = e^x$ .

$\int x \cdot e^x\,dx = x \cdot e^x - \int 1 \cdot e^x\,dx = x \cdot e^x - e^x + C$

$= e^x(x - 1) + C$

$\boxed{\int x \cdot e^x\,dx = e^x(x-1) + C}$

$\displaystyle\int x \cdot \cos(x)\,dx$

Wahl: $u = x$ , $v' = \cos(x)$ → $u' = 1$ , $v = \sin(x)$ .

$\int x \cdot \cos(x)\,dx = x \cdot \sin(x) - \int \sin(x)\,dx$

$= x\sin(x) + \cos(x) + C$

$\boxed{\int x \cdot \cos(x)\,dx = x\sin(x) + \cos(x) + C}$

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