Evaluate the integral integrate of ((x^3-4x^2+3x+1)/(x^2+4))dx
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calendarEducator since 2011
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Since the degree of numerator is larger than degree of denominator, you need to use reminder theorem such that:
`A(x) = B(x)C(x) + R(x)`
Notice that C(x) represents the quotient and R(x) represents the reminder.
`x^3-4x^2+3x+1 = (x^2+4)(ax+b) + cx + d`
`x^3-4x^2+3x+1 = ax^3 + bx^2 + 4ax + 4b + cx + d`
`x^3-4x^2+3x+1 = ax^3 + bx^2 + x(4a+c) + 4b + d`
Equating the coefficients of like powers yields:
`a = 1`
`b = -4`
`4a + c = 3 => c = -1`
`4b + d = 1 => -16 + d = 1 => d = 17`
`x^3-4x^2+3x+1 = (x^2+4)(x-4) - x + 17`
Dividing both sides by `x^2 + 4` yields:
`(x^3-4x^2+3x+1)/(x^2+4) = x - 4 + (- x + 17)/(x^2+4)`
Integrating both sides yields:
`int (x^3-4x^2+3x+1)/(x^2+4) dx= int x dx- 4 int dx+ int (- x + 17)/(x^2+4)dx`
`int (x^3-4x^2+3x+1)/(x^2+4) dx = x^2/2 - 4x - int x/(x^2+4)dx +17 int 1/(x^2+4) dx`
You should use the following substitution to solve `int x/(x^2+4) dx ` such that:
`x^2 + 4 = t => 2xdx = dt => xdx = (dt)/2`
`int x/(x^2+4) dx= (1/2)int (dt)/t = (1/2)ln|t|+c`
`int (x^3-4x^2+3x+1)/(x^2+4) dx = x^2/2 - 4x - (1/2)ln(x^2+4) + 17/2*arctan (x/2) + c`
Hence, evaluating the given integral yields `int (x^3-4x^2+3x+1)/(x^2+4) dx = x^2/2 - 4x - (1/2)ln(x^2+4) + 17/2*arctan (x/2) + c.`
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calendarEducator since 2011
write3,010 answers
starTop subjects are Math, Science, and Business
Evaluate `int (x^3-4x^2+3x+1)/(x^2+4)dx`
Rewrite the integrand using long division:
`=int (x-4+(17-x)/(x^2+4))dx`
`=int xdx-4int dx+17int(dx)/(x^2+4)-intx/(x^2+4)dx`
We will add the constant of integration at the end:
`int xdx=1/2x^2`
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`-4intdx=-4x`
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`int x/(x^2+4)` Let `u=x^2+4,du=2xdx` so we get
`int x/(x^2+4)=1/2int(du)/u=1/2ln|u|=1/2ln|x^2+4|`
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`int(dx)/(x^2+4)` is in the form `int (du)/(a^2+u^2)=1/(a) arctan(u/a)` where `a=2,u=x` so `int(dx)/(x^2+4)=1/(2) arctan(x/2)`
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All together we have:
`int(x^3-4x^2+3x+1)/(x^2+4) dx=1/2x^2 -4x-1/2ln|x^2+4|+17[1/(2) arctan(x/2)]+C`
You can factor out the i/2 to get:
`=1/2[x^2-8x-ln|x^2+4|+17 arctan(x/2)]+C`