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Here, we show you a step-by-step solved example of matrices. This solution was automatically generated by our smart calculator:
β« 1 ( x β 1 ) 2 ( x + 4 ) 2 d x \int\frac{1}{\left(x-1\right)^2\left(x+4\right)^2}dx β« ( x β 1 ) 2 ( x + 4 ) 2 1 β d x
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Passi intermedi
Rewrite the fraction 1 ( x β 1 ) 2 ( x + 4 ) 2 \frac{1}{\left(x-1\right)^2\left(x+4\right)^2} ( x β 1 ) 2 ( x + 4 ) 2 1 β 4 4 4
1 ( x β 1 ) 2 ( x + 4 ) 2 = A ( x β 1 ) 2 + B ( x + 4 ) 2 + C x β 1 + D x + 4 \frac{1}{\left(x-1\right)^2\left(x+4\right)^2}=\frac{A}{\left(x-1\right)^2}+\frac{B}{\left(x+4\right)^2}+\frac{C}{x-1}+\frac{D}{x+4} ( x β 1 ) 2 ( x + 4 ) 2 1 β = ( x β 1 ) 2 A β + ( x + 4 ) 2 B β + x β 1 C β + x + 4 D β
Find the values for the unknown coefficients: A , B , C , D A, B, C, D A , B , C , D ( x β 1 ) 2 ( x + 4 ) 2 \left(x-1\right)^2\left(x+4\right)^2 ( x β 1 ) 2 ( x + 4 ) 2
1 = ( x β 1 ) 2 ( x + 4 ) 2 ( A ( x β 1 ) 2 + B ( x + 4 ) 2 + C x β 1 + D x + 4 ) 1=\left(x-1\right)^2\left(x+4\right)^2\left(\frac{A}{\left(x-1\right)^2}+\frac{B}{\left(x+4\right)^2}+\frac{C}{x-1}+\frac{D}{x+4}\right) 1 = ( x β 1 ) 2 ( x + 4 ) 2 ( ( x β 1 ) 2 A β + ( x + 4 ) 2 B β + x β 1 C β + x + 4 D β )
1 = ( x β 1 ) 2 ( x + 4 ) 2 A ( x β 1 ) 2 + ( x β 1 ) 2 ( x + 4 ) 2 B ( x + 4 ) 2 + ( x β 1 ) 2 ( x + 4 ) 2 C x β 1 + ( x β 1 ) 2 ( x + 4 ) 2 D x + 4 1=\frac{\left(x-1\right)^2\left(x+4\right)^2A}{\left(x-1\right)^2}+\frac{\left(x-1\right)^2\left(x+4\right)^2B}{\left(x+4\right)^2}+\frac{\left(x-1\right)^2\left(x+4\right)^2C}{x-1}+\frac{\left(x-1\right)^2\left(x+4\right)^2D}{x+4} 1 = ( x β 1 ) 2 ( x β 1 ) 2 ( x + 4 ) 2 A β + ( x + 4 ) 2 ( x β 1 ) 2 ( x + 4 ) 2 B β + x β 1 ( x β 1 ) 2 ( x + 4 ) 2 C β + x + 4 ( x β 1 ) 2 ( x + 4 ) 2 D β
1 = ( x + 4 ) 2 A + ( x β 1 ) 2 B + ( x β 1 ) ( x + 4 ) 2 C + ( x β 1 ) 2 ( x + 4 ) D 1=\left(x+4\right)^2A+\left(x-1\right)^2B+\left(x-1\right)\left(x+4\right)^2C+\left(x-1\right)^2\left(x+4\right)D 1 = ( x + 4 ) 2 A + ( x β 1 ) 2 B + ( x β 1 ) ( x + 4 ) 2 C + ( x β 1 ) 2 ( x + 4 ) D
Assigning values to x x x
1 = 25 A β
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( x = 1 ) 1 = 9 A + 4 B β 18 C + 12 D β
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( x = β 1 ) 1 = 25 B β
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( x = β 4 ) 1 = 64 A + 9 B + 192 C + 72 D β
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( x = 4 ) \begin{matrix}1=25A&\:\:\:\:\:\:\:(x=1) \\ 1=9A+4B-18C+12D&\:\:\:\:\:\:\:(x=-1) \\ 1=25B&\:\:\:\:\:\:\:(x=-4) \\ 1=64A+9B+192C+72D&\:\:\:\:\:\:\:(x=4)\end{matrix} 1 = 25 A 1 = 9 A + 4 B β 18 C + 12 D 1 = 25 B 1 = 64 A + 9 B + 192 C + 72 D β ( x = 1 ) ( x = β 1 ) ( x = β 4 ) ( x = 4 ) β
Proceed to solve the system of linear equations
25 A + 0 B + 0 C + 0 D = 1 9 A + 4 B β 18 C + 12 D = 1 0 A + 25 B + 0 C + 0 D = 1 64 A + 9 B + 192 C + 72 D = 1 \begin{matrix}25A & + & 0B & + & 0C & + & 0D & =1 \\ 9A & + & 4B & - & 18C & + & 12D & =1 \\ 0A & + & 25B & + & 0C & + & 0D & =1 \\ 64A & + & 9B & + & 192C & + & 72D & =1\end{matrix} 25 A 9 A 0 A 64 A β + + + + β 0 B 4 B 25 B 9 B β + β + + β 0 C 18 C 0 C 192 C β + + + + β 0 D 12 D 0 D 72 D β = 1 = 1 = 1 = 1 β
Rewrite as a coefficient matrix
( 25 0 0 0 1 9 4 β 18 12 1 0 25 0 0 1 64 9 192 72 1 ) \left(\begin{matrix}25 & 0 & 0 & 0 & 1 \\ 9 & 4 & -18 & 12 & 1 \\ 0 & 25 & 0 & 0 & 1 \\ 64 & 9 & 192 & 72 & 1\end{matrix}\right) β 25 9 0 64 β 0 4 25 9 β 0 β 18 0 192 β 0 12 0 72 β 1 1 1 1 β β
Reducing the original matrix to a identity matrix using Gaussian Elimination
( 1 0 0 0 1 25 0 1 0 0 1 25 0 0 1 0 β 2 125 0 0 0 1 2 125 ) \left(\begin{matrix}1 & 0 & 0 & 0 & \frac{1}{25} \\ 0 & 1 & 0 & 0 & \frac{1}{25} \\ 0 & 0 & 1 & 0 & -\frac{2}{125} \\ 0 & 0 & 0 & 1 & \frac{2}{125}\end{matrix}\right) β 1 0 0 0 β 0 1 0 0 β 0 0 1 0 β 0 0 0 1 β 25 1 β 25 1 β β 125 2 β 125 2 β β β
The integral of 1 ( x β 1 ) 2 ( x + 4 ) 2 \frac{1}{\left(x-1\right)^2\left(x+4\right)^2} ( x β 1 ) 2 ( x + 4 ) 2 1 β
1 25 ( x β 1 ) 2 + 1 25 ( x + 4 ) 2 + β 2 125 ( x β 1 ) + 2 125 ( x + 4 ) \frac{1}{25\left(x-1\right)^2}+\frac{1}{25\left(x+4\right)^2}+\frac{-2}{125\left(x-1\right)}+\frac{2}{125\left(x+4\right)} 25 ( x β 1 ) 2 1 β + 25 ( x + 4 ) 2 1 β + 125 ( x β 1 ) β 2 β + 125 ( x + 4 ) 2 β
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Rewrite the fraction 1 ( x β 1 ) 2 ( x + 4 ) 2 \frac{1}{\left(x-1\right)^2\left(x+4\right)^2} ( x β 1 ) 2 ( x + 4 ) 2 1 β 4 4 4
1 25 ( x β 1 ) 2 + 1 25 ( x + 4 ) 2 + β 2 125 ( x β 1 ) + 2 125 ( x + 4 ) \frac{1}{25\left(x-1\right)^2}+\frac{1}{25\left(x+4\right)^2}+\frac{-2}{125\left(x-1\right)}+\frac{2}{125\left(x+4\right)} 25 ( x β 1 ) 2 1 β + 25 ( x + 4 ) 2 1 β + 125 ( x β 1 ) β 2 β + 125 ( x + 4 ) 2 β
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Expand the integral β« ( 1 25 ( x β 1 ) 2 + 1 25 ( x + 4 ) 2 + β 2 125 ( x β 1 ) + 2 125 ( x + 4 ) ) d x \int\left(\frac{1}{25\left(x-1\right)^2}+\frac{1}{25\left(x+4\right)^2}+\frac{-2}{125\left(x-1\right)}+\frac{2}{125\left(x+4\right)}\right)dx β« ( 25 ( x β 1 ) 2 1 β + 25 ( x + 4 ) 2 1 β + 125 ( x β 1 ) β 2 β + 125 ( x + 4 ) 2 β ) d x 4 4 4
β« 1 25 ( x β 1 ) 2 d x + β« 1 25 ( x + 4 ) 2 d x + β« β 2 125 ( x β 1 ) d x + β« 2 125 ( x + 4 ) d x \int\frac{1}{25\left(x-1\right)^2}dx+\int\frac{1}{25\left(x+4\right)^2}dx+\int\frac{-2}{125\left(x-1\right)}dx+\int\frac{2}{125\left(x+4\right)}dx β« 25 ( x β 1 ) 2 1 β d x + β« 25 ( x + 4 ) 2 1 β d x + β« 125 ( x β 1 ) β 2 β d x + β« 125 ( x + 4 ) 2 β d x
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Passi intermedi
Take the constant 1 25 \frac{1}{25} 25 1 β
1 25 β« 1 ( x β 1 ) 2 d x \frac{1}{25}\int\frac{1}{\left(x-1\right)^2}dx 25 1 β β« ( x β 1 ) 2 1 β d x
Apply the formula: β« n ( x + a ) c d x \int\frac{n}{\left(x+a\right)^c}dx β« ( x + a ) c n β d x = β n ( c β 1 ) ( x + a ) ( c β 1 ) + C =\frac{-n}{\left(c-1\right)\left(x+a\right)^{\left(c-1\right)}}+C = ( c β 1 ) ( x + a ) ( c β 1 ) β n β + C a = β 1 a=-1 a = β 1 c = 2 c=2 c = 2 n = 1 n=1 n = 1
1 25 β 1 ( 2 β 1 ) ( x β 1 ) ( 2 β 1 ) \frac{1}{25}\frac{-1}{\left(2-1\right)\left(x-1\right)^{\left(2-1\right)}} 25 1 β ( 2 β 1 ) ( x β 1 ) ( 2 β 1 ) β 1 β
β 1 25 ( x β 1 ) \frac{-1}{25\left(x-1\right)} 25 ( x β 1 ) β 1 β
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The integral β« 1 25 ( x β 1 ) 2 d x \int\frac{1}{25\left(x-1\right)^2}dx β« 25 ( x β 1 ) 2 1 β d x β 1 25 ( x β 1 ) \frac{-1}{25\left(x-1\right)} 25 ( x β 1 ) β 1 β
β 1 25 ( x β 1 ) \frac{-1}{25\left(x-1\right)} 25 ( x β 1 ) β 1 β
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Take the constant 1 25 \frac{1}{25} 25 1 β
1 25 β« 1 ( x + 4 ) 2 d x \frac{1}{25}\int\frac{1}{\left(x+4\right)^2}dx 25 1 β β« ( x + 4 ) 2 1 β d x
Apply the formula: β« n ( x + a ) c d x \int\frac{n}{\left(x+a\right)^c}dx β« ( x + a ) c n β d x = β n ( c β 1 ) ( x + a ) ( c β 1 ) + C =\frac{-n}{\left(c-1\right)\left(x+a\right)^{\left(c-1\right)}}+C = ( c β 1 ) ( x + a ) ( c β 1 ) β n β + C a = 4 a=4 a = 4 c = 2 c=2 c = 2 n = 1 n=1 n = 1
1 25 β 1 ( 2 β 1 ) ( x + 4 ) ( 2 β 1 ) \frac{1}{25}\frac{-1}{\left(2-1\right)\left(x+4\right)^{\left(2-1\right)}} 25 1 β ( 2 β 1 ) ( x + 4 ) ( 2 β 1 ) β 1 β
β 1 25 ( x + 4 ) \frac{-1}{25\left(x+4\right)} 25 ( x + 4 ) β 1 β
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The integral β« 1 25 ( x + 4 ) 2 d x \int\frac{1}{25\left(x+4\right)^2}dx β« 25 ( x + 4 ) 2 1 β d x β 1 25 ( x + 4 ) \frac{-1}{25\left(x+4\right)} 25 ( x + 4 ) β 1 β
β 1 25 ( x + 4 ) \frac{-1}{25\left(x+4\right)} 25 ( x + 4 ) β 1 β
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Take the constant 1 125 \frac{1}{125} 125 1 β
1 125 β« β 2 x β 1 d x \frac{1}{125}\int\frac{-2}{x-1}dx 125 1 β β« x β 1 β 2 β d x
Apply the formula: β« n x + b d x \int\frac{n}{x+b}dx β« x + b n β d x = n s i g n ( x ) ln β‘ ( x + b ) + C =nsign\left(x\right)\ln\left(x+b\right)+C = n s i g n ( x ) ln ( x + b ) + C b = β 1 b=-1 b = β 1 n = β 2 n=-2 n = β 2
β 2 ( 1 125 ) ln β‘ β£ x β 1 β£ -2\left(\frac{1}{125}\right)\ln\left|x-1\right| β 2 ( 125 1 β ) ln β£ x β 1 β£
Multiply the fraction and term in β 2 ( 1 125 ) ln β‘ β£ x β 1 β£ -2\left(\frac{1}{125}\right)\ln\left|x-1\right| β 2 ( 125 1 β ) ln β£ x β 1 β£
β 2 125 ln β‘ β£ x β 1 β£ -\frac{2}{125}\ln\left|x-1\right| β 125 2 β ln β£ x β 1 β£
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The integral β« β 2 125 ( x β 1 ) d x \int\frac{-2}{125\left(x-1\right)}dx β« 125 ( x β 1 ) β 2 β d x β 2 125 ln β‘ ( x β 1 ) -\frac{2}{125}\ln\left(x-1\right) β 125 2 β ln ( x β 1 )
β 2 125 ln β‘ ( x β 1 ) -\frac{2}{125}\ln\left(x-1\right) β 125 2 β ln ( x β 1 )
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Take the constant 1 125 \frac{1}{125} 125 1 β
1 125 β« 2 x + 4 d x \frac{1}{125}\int\frac{2}{x+4}dx 125 1 β β« x + 4 2 β d x
Apply the formula: β« n x + b d x \int\frac{n}{x+b}dx β« x + b n β d x = n s i g n ( x ) ln β‘ ( x + b ) + C =nsign\left(x\right)\ln\left(x+b\right)+C = n s i g n ( x ) ln ( x + b ) + C b = 4 b=4 b = 4 n = 2 n=2 n = 2
2 ( 1 125 ) ln β‘ β£ x + 4 β£ 2\left(\frac{1}{125}\right)\ln\left|x+4\right| 2 ( 125 1 β ) ln β£ x + 4 β£
Multiply the fraction and term in 2 ( 1 125 ) ln β‘ β£ x + 4 β£ 2\left(\frac{1}{125}\right)\ln\left|x+4\right| 2 ( 125 1 β ) ln β£ x + 4 β£
2 125 ln β‘ β£ x + 4 β£ \frac{2}{125}\ln\left|x+4\right| 125 2 β ln β£ x + 4 β£
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The integral β« 2 125 ( x + 4 ) d x \int\frac{2}{125\left(x+4\right)}dx β« 125 ( x + 4 ) 2 β d x 2 125 ln β‘ ( x + 4 ) \frac{2}{125}\ln\left(x+4\right) 125 2 β ln ( x + 4 )
2 125 ln β‘ ( x + 4 ) \frac{2}{125}\ln\left(x+4\right) 125 2 β ln ( x + 4 )
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Gather the results of all integrals
β 1 25 ( x β 1 ) + β 1 25 ( x + 4 ) β 2 125 ln β‘ β£ x β 1 β£ + 2 125 ln β‘ β£ x + 4 β£ \frac{-1}{25\left(x-1\right)}+\frac{-1}{25\left(x+4\right)}-\frac{2}{125}\ln\left|x-1\right|+\frac{2}{125}\ln\left|x+4\right| 25 ( x β 1 ) β 1 β + 25 ( x + 4 ) β 1 β β 125 2 β ln β£ x β 1 β£ + 125 2 β ln β£ x + 4 β£
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As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration C C C
β 1 25 ( x β 1 ) + β 1 25 ( x + 4 ) β 2 125 ln β‘ β£ x β 1 β£ + 2 125 ln β‘ β£ x + 4 β£ + C 0 \frac{-1}{25\left(x-1\right)}+\frac{-1}{25\left(x+4\right)}-\frac{2}{125}\ln\left|x-1\right|+\frac{2}{125}\ln\left|x+4\right|+C_0 25 ( x β 1 ) β 1 β + 25 ( x + 4 ) β 1 β β 125 2 β ln β£ x β 1 β£ + 125 2 β ln β£ x + 4 β£ + C 0 β
ξ Risposta finale al problema
β 1 25 ( x β 1 ) + β 1 25 ( x + 4 ) β 2 125 ln β‘ β£ x β 1 β£ + 2 125 ln β‘ β£ x + 4 β£ + C 0 \frac{-1}{25\left(x-1\right)}+\frac{-1}{25\left(x+4\right)}-\frac{2}{125}\ln\left|x-1\right|+\frac{2}{125}\ln\left|x+4\right|+C_0 25 ( x β 1 ) β 1 β + 25 ( x + 4 ) β 1 β β 125 2 β ln β£ x β 1 β£ + 125 2 β ln β£ x + 4 β£ + C 0 β ξ