Integration (1 Viewer)

excellence1 · Feb 7, 2016

Differentiate : 2X^2 +1/ 3X^2-4

Hence evaluate ∫ between the limits 1 and 0, x/(3x^2 -4)^2

davidgoes4wce · Feb 7, 2016

Learning from my lesson last night, about not using substitution.

Nailgun · Feb 7, 2016

$y=\frac { 2{ x }^{ 2 }+1 }{ { 3 }x^{ 2 }-4 } \\ \frac { dy }{ dx } =\frac { vu'-uv' }{ v^{ 2 } } =\frac { 4x(3{ x }^{ 2 }-4)-6x(2{ x }^{ 2 }+1) }{ (3{ x }^{ 2 }-4)^{ 2 } } =\frac { -22x }{ (3{ x }^{ 2 }-4)^{ 2 } } \\ Hence,\quad \int _{ 0 }^{ 1 }{ \frac { x }{ (3{ x }^{ 2 }-4)^{ 2 } } } dx=\left[ \frac { 2{ x }^{ 2 }+1 }{ { 3 }x^{ 2 }-4 } .\frac { 1 }{ -22 } \right] _{ 0 }^{ 1 }=\left[ -\frac { 2{ x }^{ 2 }+1 }{ 22({ 3 }x^{ 2 }-4) } \right] _{ 0 }^{ 1 }\\ \int _{ 0 }^{ 1 }{ \frac { x }{ (3{ x }^{ 2 }-4)^{ 2 } } } dx=\frac { 3 }{ 22 } -\frac { 1 }{ 88 } =\frac { 1 }{ 8 } { u }^{ 2 }$

edit: whoops too slow

excellence1 · Feb 7, 2016

davidgoes4wce said:
Learning from my lesson last night, about not using substitution.

What's life without mistakes, thanks for your help and thanks Nailgun

Integration (1 Viewer)

excellence1

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