Find β« ln 3π₯ β 1 π₯ 5 + cos 4π₯ β π βπ₯ 2 β πx
β«ln(3x)βx5+cos(4x)βeβx2\int ln(3x) -x^5 +cos(4x) -e^{\dfrac{-x}{2}}β«ln(3x)βx5+cos(4x)βe2βxβ
β«1Γln(3x)βx5+cos(4x)βeβx2\int 1\times ln(3x) -x^5 +cos(4x) -e^{\dfrac{-x}{2}}β«1Γln(3x)βx5+cos(4x)βe2βxβ
[ln(3x)xββ«xΓ1x]+sin(4x)4βx66β(β2eβx2)[ln(3x)x-\int x \times \dfrac{1}{x} ] +\dfrac{sin(4x)}{4}-{ \dfrac{x^6}{6} } -(-2e^{\dfrac{-x}{2}})[ln(3x)xββ«xΓx1β]+4sin(4x)ββ6x6ββ(β2e2βxβ)
xln(3x)βx+sin(4x)4βx66+2eβx2+Cxln(3x)- x +\dfrac{sin(4x)}{4}-{ \dfrac{x^6}{6} } +2e^{\dfrac{-x}{2}} +Cxln(3x)βx+4sin(4x)ββ6x6β+2e2βxβ+C
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