Question

Va rog sa ma ajutati!

1. $\int\limits {xe^{-x} } \, dx$

2. $\int\limits {x*a^x} \, dx$ , x ∈ R, a ∈ (0,+∞) \ {1}

3. $\int\limits {(x^2+x+1)lnx} \, dx$ , x∈ (0,+∞), a∈ R*

Answer 1

$1)\int x {e}^{ - x} \: dx$

$f = x = > f' = x' = 1$

$g' = {e}^{ - x} = > g = \int {e}^{ - x} \: dx = - {e}^{ - x}$

$\int x {e}^{ - x}\:dx = x \times ( - {e}^{ - x} ) - \int - {e}^{ - x} \times 1 \: dx$

$= - x {e}^{ - x} - \int - {e}^{ - x} \: dx$

$= - x {e}^{ - x} + \int {e}^{ - x} \: dx$

$= - x {e}^{ - x} - {e}^{ - x}$

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

$2) \int x {a}^{x} \: dx$

$f = x = > f' = x' = 1$

$g' = {a}^{x} = > g = \int {a}^{x} \: dx = \frac{ {a}^{x} }{lna}$

$\int x {a}^{x} \: dx= \frac{x {a}^{x} }{lna} - \int \frac{ {a}^{x} }{lna} \: dx$

$= \frac{x {a}^{x} }{lna} - \frac{1}{lna} \int {a}^{x} \: dx$

$= \frac{x {a}^{x} }{lna} - \frac{1}{lna} \times \frac{ {a}^{x} }{lna}$

$= \frac{x {a}^{x} }{lna} - \frac{ {a}^{x} }{ {ln}^{2} a} = \frac{x {a}^{x} lna - {a}^{x} }{ {ln}^{2}a } = \frac{ {a}^{x} (xlna - 1)}{ {ln}^{2}a } + C$

$3) \int( {x}^{2} + x + 1)lnx \: dx$

$f = lnx = > f' = (lnx)' = \frac{1}{x}$

$g' = {x}^{2} + x + 1$

$g = \int ({x}^{2} + x + 1) \: dx$

$g = \int {x}^{2} \: dx + \int x \: dx + \int 1 \: dx$

$g = \frac{ {x}^{2 + 1} }{2 + 1} + \frac{ {x}^{1 + 1} }{1 + 1} + x$

$g = \frac{ {x}^{3} }{3} + \frac{ {x}^{2} }{2} + x$

$\int( {x}^{2} + x + 1)lnx\:dx$

$= lnx( \frac{ {x}^{3} }{3} + \frac{ {x}^{2} }{2} + x) - \int \frac{ \frac{ {x}^{3} }{3} + \frac{ {x}^{2} }{2} + x }{x} \: dx$

$\int \frac{ \frac{ {x}^{3} }{3} + \frac{ {x}^{2} }{2} + x }{x} \: dx$

$= \int {x}^{ - 1} ( \frac{ {x}^{3} }{3} + \frac{ {x}^{2} }{2} + x) \: dx$

$= \int \frac{ {x}^{ - 1} \times {x}^{3} }{3} + \frac{ {x}^{ - 1} \times {x}^{2} }{2} + {x}^{ - 1} \times x \: dx$

$= \int \frac{ {x}^{ - 1 + 3} }{3} + \frac{ {x}^{ - 1 + 2} }{2} + {x}^{ - 1 + 1} \: dx$

$= \int \frac{ {x}^{2} }{3} + \frac{x}{2} + 1 \: dx$

$\int (\frac{ {x}^{2} }{3} + \frac{x}{2} + 1 )\: dx$

$= \int \frac{ {x}^{2} }{3} \: dx + \int \frac{x}{2} \: dx + \int1 \: dx$

$= \frac{1}{3} \int {x}^{2} \: dx + \frac{1}{2} \int x \:dx+ x$

$= \frac{1}{3} \times \frac{ {x}^{2 + 1} }{2 + 1} + \frac{1}{2} \times \frac{ {x}^{1 + 1} }{1 + 1} + x$

$= \frac{1}{3} \times \frac{ {x}^{3} }{3} + \frac{1}{2} \times \frac{ {x}^{2} }{2} + x$

$= \frac{ {x}^{3} }{9} + \frac{ {x}^{2} }{4} + x + C$

$\int( {x}^{2} + x + 1)lnx \: dx$

$= lnx( \frac{ {x}^{3} }{3} + \frac{ {x}^{2} }{2} + x) - ( \frac{ {x}^{3} }{9} + \frac{ {x}^{2} }{4} + x)$

$= lnx( \frac{ {x}^{3} }{3} + \frac{ {x}^{2} }{2} + x) - \frac{ {x}^{3} }{9} - \frac{ {x}^{2} }{4} - x + C$