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int(x+4/x^2-5x+6)

Guest Jun 16, 2017
 #1
avatar+7155 
+2

\(\int(x+\frac4{x^2}-5x+6)\,dx \\~\\ =\int(\frac4{x^2}-4x+6)\,dx \\~\\ =\int \frac4{x^2}\,dx+\int-4x\,dx+\int6\,dx \\~\\ =4\int x^{-2}\,dx-4\int x\,dx+6\int 1\,dx\)       Combine like terms.

 

Applying the power rule gives us...

 

\(=4(\frac{x^{-1}}{-1})-4(\frac{x^2}{2})+6(x) +c\\~\\ =-4x^{-1}-2x^2+6x+c\)                 , where   c   is a constant.

hectictar  Jun 16, 2017
 #2
avatar+19653 
+1

integral( (x+4) / (x^2-5x+6) ) dx

 

\(\begin{array}{rcl} && \int { \frac{x+4} {x^2-5x+6} \ dx} \quad & | \quad x^2-5x+6 = (x-3)(x-2) \\ &=& \int { \frac{x+4} {(x-3)(x-2)} \ dx} \\ \end{array} \)

 

Partial fraction decomposition:

\(\begin{array}{|lrcll|} \hline & \frac{x+4} {(x-3)(x-2)} &=& \frac{A}{x-3} + \frac{B}{x-2} \\ & x+4 &=& A\cdot (x-2) + B\cdot (x-3) \\ x = 2: & 2+4 &=& A\cdot (2-2) + B\cdot (2-3) \\ & 6 &=& 0 + B\cdot (-1) \\ & 6 &=& B\cdot (-1) \\ & \mathbf{B} &\mathbf{=}& \mathbf{-6} \\ \\ x = 3: & 3+4 &=& A\cdot (3-2) + B\cdot (3-3) \\ & 7 &=& A\cdot (1) + B\cdot (0) \\ & 7 &=& A\cdot (1) \\ & \mathbf{A} &\mathbf{=}& \mathbf{7} \\\\ & \mathbf{\frac{x+4} {(x-3)(x-2)}} &\mathbf{=}& \mathbf{\frac{7}{x-3} - \frac{6}{x-2}} \\ \hline \end{array} \)

 

\(\begin{array}{rcl} && \mathbf{ \int { \frac{x+4} {x^2-5x+6} \ dx} } \\ &=& \int { \frac{x+4} {(x-3)(x-2)} \ dx} \\ &=& \int { \Big(\frac{7}{x-3} - \frac{6}{x-2} \Big) \ dx} \\ &=& 7\cdot \int { \frac{1}{x-3} \ dx} -6\int { \frac{1}{x-2} \ dx} \\ &\mathbf{=}& \mathbf{7\cdot \ln(|x-3|) -6\cdot \ln(|x-2|) +c }\\ \end{array} \)

 

laugh

heureka  Jun 16, 2017
edited by heureka  Jun 16, 2017
 #3
avatar
0

Take the integral:
 integral(4/x^2 - 4 x + 6) dx


Integrate the sum term by term and factor out constants:
 = -4 integral x dx + 4 integral1/x^2 dx + 6 integral1 dx


The integral of x is x^2/2:
 = -2 x^2 + 4 integral1/x^2 dx + 6 integral1 dx


The integral of 1/x^2 is -1/x:
 = -4/x - 2 x^2 + 6 integral1 dx


The integral of 1 is x:
Answer: | = -2 x^2 + 6 x - 4/x + constant

Guest Jun 16, 2017
 #4
avatar
0

As "heureka" read it:
Take the integral:
 integral(x + 4)/(x^2 - 5 x + 6) dx
Rewrite the integrand (x + 4)/(x^2 - 5 x + 6) as (2 x - 5)/(2 (x^2 - 5 x + 6)) + 13/(2 (x^2 - 5 x + 6)):
 = integral((2 x - 5)/(2 (x^2 - 5 x + 6)) + 13/(2 (x^2 - 5 x + 6))) dx
Integrate the sum term by term and factor out constants:
 = 1/2 integral(2 x - 5)/(x^2 - 5 x + 6) dx + 13/2 integral1/(x^2 - 5 x + 6) dx
For the integrand (2 x - 5)/(x^2 - 5 x + 6), substitute u = x^2 - 5 x + 6 and du = (2 x - 5) dx:
 = 1/2 integral1/u du + 13/2 integral1/(x^2 - 5 x + 6) dx
The integral of 1/u is log(u):
 = (log(u))/2 + 13/2 integral1/(x^2 - 5 x + 6) dx
For the integrand 1/(x^2 - 5 x + 6), complete the square:
 = (log(u))/2 + 13/2 integral1/((x - 5/2)^2 - 1/4) dx
For the integrand 1/((x - 5/2)^2 - 1/4), substitute s = x - 5/2 and ds = dx:
 = (log(u))/2 + 13/2 integral1/(s^2 - 1/4) ds
Factor -1/4 from the denominator:
 = (log(u))/2 + 13/2 integral4/(4 s^2 - 1) ds
Factor out constants:
 = (log(u))/2 + 26 integral1/(4 s^2 - 1) ds
Factor -1 from the denominator:
 = (log(u))/2 - 26 integral1/(1 - 4 s^2) ds
For the integrand 1/(1 - 4 s^2), substitute p = 2 s and dp = 2 ds:
 = (log(u))/2 - 13 integral1/(1 - p^2) dp
The integral of 1/(1 - p^2) is tanh^(-1)(p):
 = (log(u))/2 - 13 tanh^(-1)(p) + constant
Substitute back for p = 2 s:
 = (log(u))/2 - 13 tanh^(-1)(2 s) + constant
Substitute back for s = x - 5/2:
 = (log(u))/2 + 13 tanh^(-1)(5 - 2 x) + constant
Substitute back for u = x^2 - 5 x + 6:
 = 1/2 log(x^2 - 5 x + 6) + 13 tanh^(-1)(5 - 2 x) + constant
Factor the answer a different way:
 = 1/2 (log(x^2 - 5 x + 6) + 26 tanh^(-1)(5 - 2 x)) + constant
Which is equivalent for restricted x values to:
Answer: | = 7 log(3 - x) - 6 log(2 - x) + constant

Guest Jun 16, 2017

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