# int (x e^(2x))/(1 + 2x)^2 dx Evaluate the integral

int(xe^(2x))/(1+2x)^2dx

If f(x) and g(x) are differentiable functions, then

intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx

If we rewrite f(x)=u and g'(x)=v, then

intuvdx=uintvdx-int(u'intvdx)dx

Using the above method of integration by parts,

Let u=xe^(2x)

u'=xd/dx(e^(2x))+e^(2x)d/dx(x)

u'=x(2e^(2x))+e^(2x)

u'=e^(2x)(2x+1)

v=1/(1+2x)^2

intvdx=int(1/(1+2x)^2)dx

Let's integrate by the use of substitution method,

Let t=1+2x

dt=2dx

int(1/(1+2x)^2)dx=intdt/(2t^2)

=1/2(t^(-2+1)/(-2+1))

=-1/(2t)

substitute back t=1+2x,

=-1/(2(1+2x))

...

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int(xe^(2x))/(1+2x)^2dx

If f(x) and g(x) are differentiable functions, then

intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx

If we rewrite f(x)=u and g'(x)=v, then

intuvdx=uintvdx-int(u'intvdx)dx

Using the above method of integration by parts,

Let u=xe^(2x)

u'=xd/dx(e^(2x))+e^(2x)d/dx(x)

u'=x(2e^(2x))+e^(2x)

u'=e^(2x)(2x+1)

v=1/(1+2x)^2

intvdx=int(1/(1+2x)^2)dx

Let's integrate by the use of substitution method,

Let t=1+2x

dt=2dx

int(1/(1+2x)^2)dx=intdt/(2t^2)

=1/2(t^(-2+1)/(-2+1))

=-1/(2t)

substitute back t=1+2x,

=-1/(2(1+2x))

int(xe^(2x))/(1+2x)^2dx=xe^(2x)int(1/(1+2x)^2)dx-int(d/dx(xe^(2x))int(1/(1+2x)^2)dx)dx)

=xe^(2x)(-1/(2(1+2x)))-inte^(2x)(1+2x)(-1/(2(1+2x)))dx

=(-xe^(2x))/(2(1+2x))+inte^(2x)/2dx

=(-xe^(2x))/(2(1+2x))+(1/2)e^(2x)/2

=e^(2x)/4-(xe^(2x))/(2(1+2x))

Add a constant C to the solution,

int(xe^(2x))/(1+2x)^2dx=e^(2x)/4-(xe^(2x))/(2(1+2x))+C

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