Question

Вычислить интегралы:
1) $\int\limits^e_1 {(x+1)} \, lnxdx$
2) $\int\limits^e_1 {xln} \, xdx$
3) $\int\limits^p_0 {xcos} \, xdx$

Answer 1

Ответ:

1.

По частям

$\int\limits^{ e } _ {1} (x + 1) ln(x) dx$

$u = ln(x) \: \: du = \frac{1}{x} dx \\ dv =( x + 1)dx \: \: v = \frac{ {x}^{2} }{2} + x \\ \\ uv - \int\limits \: vdu = \\ =( \frac{ {x}^{2} }{2} + x) ln(x) | ^{e} _ {1} - \int\limits^{ e } _ {1} \frac{1}{x} ( \frac{ {x}^{2} }{2} + x)dx = \\ = ( \frac{ {x}^{2} }{2} + x)ln(x) | ^{e} _ {1} - \int\limits^{ e } _ {1} ( \frac{x}{2} + 1)dx = \\ = (( \frac{ {x}^{2} }{2} + x) ln(x) - \frac{ {x}^{2} }{4} - x)| ^{e} _ {1} = \\ = ( \frac{ {e}^{2} }{2} + e) ln(e) - \frac{ {e}^{2} }{4} - e - ( \frac{1}{2} + 1) ln(1) - \frac{1}{4} - 1) = \\ = \frac{ {e}^{2} }{2} + e - \frac{ {e}^{2} }{4} - e + \frac{5}{4} = \frac{e {}^{2} }{4} + \frac{5}{4} = \frac{1}{4} ( {e}^{2} + 5)$

2.

$\int\limits^{ e } _ {1} x ln(x) dx$

$u = ln(x) \: \: \: \: \: du = \frac{1}{x} dx \\ dv = xdx \: \: \: \: \: dv = \frac{ {x}^{2} }{2} \\ \\ \frac{ {x}^{2} }{2} ln(x) | ^{ e } _ {1} - \int\limits^{ e } _ {1} \frac{1}{x} \times \frac{ {x}^{2} }{2} dx = \\ = \frac{ {x}^{2} }{2} ln(x) | ^{ e } _ {1} - \int\limits^{ e } _ {1} \frac{x}{2} dx = \\ = ( \frac{ {x}^{2} }{2} ln(x) - \frac{ {x}^{2} }{4} )| ^{ e } _ {1} = \\ = \frac{ {e}^{2} }{2} ln(e) - \frac{ {e}^{2} }{4} - \frac{1}{2} \times 0 + \frac{1}{4} = \\ = \frac{ {e}^{2} }{4} + \frac{1}{4} = \frac{1}{4} ( {e}^{2} + 1)$

3.

$\int\limits^{ \pi} _ {0} x \cos(x) dx \\ \\ \\ u = x \: \: \: \: \: \: \: \: du = dx \\ dv \cos(x)dx \: \: \: \: \: \: v = \sin(x) \\ \\ x \sin(x) | ^{ \pi } _ {0} - \int\limits^{ \pi} _ {0} \sin(x) dx = \\ = (x \sin(x) + \cos(x) ) | ^{ \pi} _ {0} = \\ = \pi \sin(\pi) + \cos(\pi) - 0 - \cos(0) = \\ =0-1-1=-2$