Qual é a integral de $\int [\frac{x e^{2 x}}{{(2 x + 1)}^{2}}] d x$ $int [(xe ^ (2x)) / (2x + 1) ^ 2] dx$ ?

Outro método:

$I = \int \frac{x e^{2 x}}{{(2 x + 1)}^{2}} d x$ $I=int(xe^(2x))/(2x+1)^2dx$

Nós podemos tentar Integração por partes com $u = e^{2 x}$ $u=e^(2x)$ e $d v = \frac{x}{{(2 x + 1)}^{2}} d x$ $dv=x/(2x+1)^2dx$ .

Observe que $v = \int \frac{x}{{(2 x + 1)}^{2}} d x$ $v=intx/(2x+1)^2dx$ . De locação $t = 2 x + 1$ $t=2x+1$ , isso implica que $x = \frac{1}{2} (t - 1)$ $x=1/2(t-1)$ e que $d t = 2 d x \Rightarrow d x = \frac{1}{2} d t$ $dt=2dx=>dx=1/2dt$ , assim $v = \int \frac{\frac{1}{2} (t - 1)}{t^{2}} \frac{1}{2} d t = \frac{1}{4} \int (\frac{1}{t} - \frac{1}{t^{2}}) d t = \frac{1}{4} ln | t | + \frac{1}{4 t} ...$ $v=int(1/2(t-1))/t^2 1/2dt=1/4int(1/t-1/t^2)dt=1/4lnabst+1/(4t)...$

Tb, $d u = 2 e^{2 x} d x$ $du=2e^(2x)dx$ .

Então:

$I = \frac{1}{4} e^{2 x} (ln | 2 x + 1 | + \frac{1}{2 x + 1}) - \int 2 e^{2 x} \frac{1}{4} (ln | 2 x + 1 | + \frac{1}{2 x + 1}) d x$ $I=1/4e^(2x)(lnabs(2x+1)+1/(2x+1))-int2e^(2x)1/4(lnabs(2x+1)+1/(2x+1))dx$

$I = \frac{e^{2 x}}{4} ln | 2 x + 1 | + \frac{e^{2 x}}{4 (2 x + 1)} - \frac{1}{2} \int e^{2 x} ln | 2 x + 1 | d x - \frac{1}{2} \int \frac{e^{2 x}}{2 x + 1} d x$ $I=e^(2x)/4lnabs(2x+1)+e^(2x)/(4(2x+1))-1/2inte^(2x)lnabs(2x+1)dx-1/2inte^(2x)/(2x+1)dx$

Tentando IBP na segunda integral, deixe:

$u = - \frac{1}{2} e^{2 x} \Rightarrow d u = - e^{2 x} d x$ $u=-1/2e^(2x)=>du=-e^(2x)dx$
$d v = \frac{d x}{2 x + 1} \Rightarrow v = \frac{1}{2} ln | 2 x + 1 |$ $dv=dx/(2x+1)=>v=1/2lnabs(2x+1)$

Assim:

$I = \frac{e^{2 x}}{4} ln | 2 x + 1 | + \frac{e^{2 x}}{4 (2 x + 1)} - \frac{1}{2} \int e^{2 x} ln | 2 x + 1 | d x - \frac{e^{2 x}}{4} ln | 2 x + 1 | + \frac{1}{2} \int e^{2 x} ln | 2 x + 1 | d x$ $I=e^(2x)/4lnabs(2x+1)+e^(2x)/(4(2x+1))-1/2inte^(2x)lnabs(2x+1)dx-e^(2x)/4lnabs(2x+1)+1/2inte^(2x)lnabs(2x+1)dx$

$I = \frac{e^{2 x}}{4 (2 x + 1)} + C$ $I=e^(2x)/(4(2x+1))+C$

Qual é a integral de int [(xe ^ (2x)) / (2x + 1) ^ 2] dx ∫[xe2x(2x+1)2]dxint [(xe ^ (2x)) / (2x + 1) ^ 2] dx ?

Qual é a integral de $\int [\frac{x e^{2 x}}{{(2 x + 1)}^{2}}] d x$ $int [(xe ^ (2x)) / (2x + 1) ^ 2] dx$ ?