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`int_1^(+oo) x/(x+1)^2 dx`
Decompose the integrand into its equivalent partial fractions.
`x/(x+1)^2 = A/(x+1)+B/(x+1)^2`
Multiply both sides by the LCD `(x+1)^2` .
`x = A(x+1) + B`
To solve for B, set x=-1.
`-1=A(-1+1)+B ` `-1=A(0)+B` `B=-1`
To solve for A set x=0 and substitute B=-1.
`0=A(0+1)+(-1)` `0=A-1` `A=1`
So we have,
`int_1^(x) x/(x+1)^2dx =int_1^(+oo) (1/(x+1)-1/(x+1)^2) dx = int_1^(+oo) 1/(x+1)dx - int_1^(+oo)1/(x+1)^2dx`
Since the upper limit of the integral is infinity, replace the infinity with t variable to be able to evaluate the integral.
`= int_1^t 1/(x+1)dx - int_1^t 1/(x+1)^2dx`
Then, let's try to evaluate the two integrals separately.
For the first integral, apply the logarithm formula which is `int1/udu=lnu+C` .
`int_1^t 1/(x+1)dx = ln(x+1)` `|_1^t = ln (t+1) - ln 2`
Note that t represents +infinity, so take the limit of t as it approaches +infinity, to evaluate further.
To do so, apply the rule for logarithm which is `lim_(x->oo) lnx= oo`
`lim_(t->+oo)` `[ln(t+1) - ln2] = oo-ln2 = oo`
The limit is infinity because if we subtract a large number with a small one, the resulting value is a large number.
Next, let's evaluate the other integral using u-substitution.
`int_1^t 1/(x+1)^2 dx`
`u = x+1` and `du=dx`
`int 1/u^2 du = int u^(-2)du= u^(-1) = 1/u`
Then, substitute back u=x+1.
`int_1^t 1/(x+1)^2 dx = 1/(x+1)` `| _1^t = 1/(t+1) - 1/(1+1) = 1/(t+1)-1/2`
Then, take the limit of t as it approaches +infinity. Note that` lim_(x->oo) 1/x = 0` .
`lim_(t->+oo)` `[1/(t+1) - 1/2 ]= 0- 1/2 = -1/2`
Next substitute the limits of the two integral. So we have,
`int_1^(+oo) x/(x+1)^2 dx = int_1^(+oo) 1/(x+1) dx - int_1^(+oo)1/(x+1)^dx = oo + 1/2 = oo`
Note that if a small number is added to a large number, the sum has a large value.
Hence, `int_1^oo x/(x+1)^2dx = oo` .
Since we have infinity, the integral diverges.
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