怀疑题目有错,为-3∫[1,-1]√(4x^2+5)dx,解如下:
-3∫[1,-1]√(4x^2+5)dx
=3∫[-1,1]√(4x^2+5)dx
=6∫[0,1]√(4x^2+5)dx
=6√5∫[0,1]√[2x/√5)^2+1]dx
令2x/√5=tan t dx=√5/2sec^2tdt
-3∫[1,-1]√(4x^2+5)dx
=6√5*√5/2∫[0,arctan(2/√5)]sec^3tdt
=15∫[0,arctan(2/√5)]dt/cos^3t
=15/2 sint/cos^2t|[0,arctan(2/√5)]+15*1/2∫[0,arctan(2/√5)]dt/cost
=15/2 sint/cos^2t|[0,arctan(2/√5)]+15*1/2ln|sect+tant|[0,arctan(2/√5)]
=9+15/4*ln5
如果认定题目没错,那更简单,解如下:
-3∫[1,-1]√(4u^2+5)dx
=3∫[-1,1]√(4u^2+5)dx
=3√(4u^2+5)∫[-1,1]dx
=6√(4u^2+5)
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