请帮忙做几道数学题,用第一换元积分法求下面几个的不定积分,麻烦写一下过程
3个回答
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9)令u=√5/7*x
11)原式=3∫1/(5x^2+7)dx- 1/5∫10x/(5x^2+7)dx
前一个令u=√5/7*x , 后一个是-1/5ln|5x^2+7|
12)原式=∫x/(1+4x^2)dx - ∫√arctan2x/[1+(2x)^2]dx
=1/8∫8x/(1+4x^2)dx - 1/2∫√arctan2xdarctan2x
=1/8ln|1+4x^2|- 1/2*2/3 *(arctan2x)^(3/2)+c
=1/8ln|1+4x^2|- 1/3*(arctan2x)^(3/2)+c
13)原式=∫(sec3x)^2dx+∫sec3xtan3xdx
=1/3tan3x+1/3ln|sec3x+tan3x|+c
16)令2^x=t , x=log2 t , dx=1/(tln2)dt
原式=∫(1/[(t+1)tln2]dt = 1/ln2*∫[1/t-1/(t+1)]dt
= 1/ln2*(ln|t|-ln|t+1| )+c
= 1/ln2*(ln|2^x|-ln|2^x+1| )+c
19)令u=arctan√x,du=1/[2 (1+x)√x]dx
1/[ (1+x)√x]dx = 2du
原式=2∫udu=u^2+c
= (arctan√x)^2+c
11)原式=3∫1/(5x^2+7)dx- 1/5∫10x/(5x^2+7)dx
前一个令u=√5/7*x , 后一个是-1/5ln|5x^2+7|
12)原式=∫x/(1+4x^2)dx - ∫√arctan2x/[1+(2x)^2]dx
=1/8∫8x/(1+4x^2)dx - 1/2∫√arctan2xdarctan2x
=1/8ln|1+4x^2|- 1/2*2/3 *(arctan2x)^(3/2)+c
=1/8ln|1+4x^2|- 1/3*(arctan2x)^(3/2)+c
13)原式=∫(sec3x)^2dx+∫sec3xtan3xdx
=1/3tan3x+1/3ln|sec3x+tan3x|+c
16)令2^x=t , x=log2 t , dx=1/(tln2)dt
原式=∫(1/[(t+1)tln2]dt = 1/ln2*∫[1/t-1/(t+1)]dt
= 1/ln2*(ln|t|-ln|t+1| )+c
= 1/ln2*(ln|2^x|-ln|2^x+1| )+c
19)令u=arctan√x,du=1/[2 (1+x)√x]dx
1/[ (1+x)√x]dx = 2du
原式=2∫udu=u^2+c
= (arctan√x)^2+c
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