How do you integrate $\int \frac{6 x + 5}{{(x + 2)}^{4}}$ $int (6x+5) / (x+2) ^4$ using partial fractions?

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How do you integrate $\int \frac{6 x + 5}{{(x + 2)}^{4}}$ $int (6x+5) / (x+2) ^4$ using partial fractions?

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Answer 1 · 2016-03-30T18:53:03

$\int \frac{6 x + 5}{{(x + 2)}^{4}} d x = \int (\frac{6}{{(x + 2)}^{3}} - \frac{7}{{(x + 2)}^{4}}) d x = - \frac{2}{{(x + 2)}^{2}} + \frac{7}{3 {(x + 2)}^{3}} + C$ $int(6x+5)/(x+2)^4 dx = int (6/(x+2)^3-7/(x+2)^4)dx = -2/(x+2)^2+7/(3(x+2)^3) + C$

Explanation:

$\frac{6 x + 5}{{(x + 2)}^{4}} = \frac{A}{(x + 2)} + \frac{B}{{(x + 2)}^{2}} + \frac{C}{{(x + 2)}^{3}} + \frac{D}{{(x + 2)}^{4}}$ $(6x+5)/(x+2)^4 = A/((x+2)) + B/(x+2)^2 +C/(x+2)^3 +D/(x+2)^4$

$A {(x + 2)}^{3} + B {(x + 2)}^{2} + C (x + 2) + D = 6 x + 5$ $A(x+2)^3 + B(x+2)^2 +C(x+2) + D = 6x+5$

$A x^{3} = 0 x^{3} \Rightarrow A = 0$ $Ax^3 = 0x^3 rArr A=0$

$B x^{2} = 0 x^{2} \Rightarrow B = 0$ $Bx^2 = 0x^2 rArr B=0$

$C x = 6 x \Rightarrow C = 6$ $Cx = 6x rArr C=6$

$2 C + D = 5 and C = 6 \Rightarrow D = - 7$ $2C+D=5 and C=6 rArr D=-7$

Now evaluate $\int (6 {(x + 2)}^{- 3} - 7 {(x + 2)}^{- 4}) d x$ $int(6(x+2)^-3 - 7(x+2)^-4) dx$ to get

$- \frac{2}{{(x + 2)}^{2}} + \frac{7}{3 {(x + 2)}^{3}} + C$ $-2/(x+2)^2+7/(3(x+2)^3) + C$

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How do you integrate $\int \frac{6 x + 5}{{(x + 2)}^{4}}$ $int (6x+5) / (x+2) ^4$ using partial fractions?

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How do you integrate int (6x+5) / (x+2) ^4∫6x+5(x+2)4int (6x+5) / (x+2) ^4 using partial fractions?

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How do you integrate $\int \frac{6 x + 5}{{(x + 2)}^{4}}$ $int (6x+5) / (x+2) ^4$ using partial fractions?