I need to know how to integrate this

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I need to know how to integrate this

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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use partial fractions here\[\frac{3x+4}{(x^2+4)(x-3)}=\frac{Ax+B}{x^2+4}+\frac{C}{x-3}\]and find A,B and C
Split using partial fractions: \[\frac{ 3x+4 }{ (x ^{2}+4)(x-3) } = \frac{ A }{ x-3 } + \frac{ Bx + C}{ (x ^{2} + 4) } = \frac{ A(x^{2} +4)+(Bx+C)(x-3) }{ (x ^{2}+4)(x-3) }\] Set equal to 3, looking at just numerator: \[3*3 + 4 = 13 = A(9+4) + (3B+C)(0) = 13A \rightarrow A = 1\] Set x equal to 0: \[0 + 4 = 4 = 4A + C \rightarrow 4 = 4+C \rightarrow C = 0\] Set x equal to 1: \[3 + 4 =A(1+4) + (B+C)(1-3) = 5A -2B -2C \rightarrow 5 - 2B - 0 \rightarrow 2 = -2B \rightarrow B = -1\] Therefore: \[\frac{ 3x+4 }{ (x ^{2}+4)(x-3)} = \frac{ 1 }{ x-3 } - \frac{ x }{ x ^{2} + 4 }\] Now integrate those.

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thank you very much, you have no idea how much i love you xD
No problemo The integral of 1/(x-3) is ln(x-3) and the integral of -x/(x^2 + 4): u-sub: u = x^2 + 4, du = 2x dx \[\frac{ -1 }{ 2 } \int\limits_{}^{}\frac{ 1 }{ u }du = \frac{ -1 }{ 2 }\ln(u) = \frac{ -1 }{ 2 }\ln(x ^{2}+4)\] \[\left( \ln(2-3) - \frac{ 1 }{ 2 }\ln(4+4) \right)-\left( \ln(0-3) - \frac{ 1 }{ 2 }\ln(0+4) \right) \] \[= \ln(-1)-\frac{ 1 }{ 2 }\ln(8)-\ln(-3)-\frac{ 1 }{ 2 }\ln(4) = \ln(\frac{ 1 }{ 3\sqrt{2} })\]

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