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Question Number 73668 by liki last updated on 14/Nov/19
Qn.∫(e2x2+2x+6)dx...Ineedhelpplz..
Answered by arkanmath7@gmail.com last updated on 14/Nov/19
=∫e2x2+2x.e6dx=e6∫e2x2+2xdx=e6∫e2x2+2x+12−12dx=e6∫e(2x+12)2−12dx=e−12e6∫e(2x+12)2dxnowletu=2x+12⇒du=2dx⇒du=2dx⇒dx=12du∫e2x2+2x+6dx=e112.12∫eu2du=πe11222∫2eu2πduerrorfunction=πe11222erf(u)=πe11222erf(2x+12)+c
Commented by liki last updated on 14/Nov/19
Thankssomuch..
well soln
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