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Indefinite integrals Qs
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1a. (d/dx)(x^5lnx) { (g×k)'=g'×k+g×k' } =lnx?(d/dx)x^5 + x^5?(d/dx)lnx =5x^4?lnx+x^5?(1/x) =5x^4?lnx+x^4 1b. ∫x^4lnxdx =∫lnxd(x^5/5) =(1/5)∫lnxdx^5 =(1/5){x^5lnx-∫x^5dlnx } =(1/5){x^5lnx-∫x^5/xdx } =(1/5){x^5lnx-∫x^4dx } =(1/5){x^5lnx-x^5/5}+C =(1/25)(5x^5lnx-x^5)+C 2a. (d/dx)(2∫x^4 dx) =(d/dx)[(2/5)x^5+C] =(10/5)x^4 =2x^4 2b. ∫xlnxdx =∫lnxd(x^2/2) =(1/2)∫lnxdx^2 =(1/2)[x^2lnx-∫x^2dlnx] =(1/2)[x^2lnx-∫xdx] =(1/2)[x^2lnx-x^2/2]+C =(1/4)(2x^2lnx-x^2)+C 3. ∫(lnx+1)/2x dx =(1/2)∫(lnx+1)/x dx =(1/2)∫(lnx+1)dlnx =(1/2)[(lnx)^2/2+lnx)+C =(lnx/4)(lnx+2)+C 4. ∫x[(2x+7)^(1/2)-x] dx ( u=2x+7 → x=(u-7)/2 →du=2dx =(1/2)∫[(u-7)/2][√(u)-(u-7)/2] 2dx =(1/2)∫[(u-7)/2][√(u)-(u-7)/2]du =(1/4)∫[u^(3/2)-7u^(1/2)-(u^2-14u+49)/2]du =(1/8)∫[2u^(3/2)-14u^(1/2)-(u^2-14u+49)]du =(1/8)∫[2u^(3/2)-14u^(1/2)-u^2+14u-49]du =(1/8)[(4/5)u^(5/2)-(28/3)u^(3/2)-u^3/3+7u^2-49u]+C 自己代u= = 5. ∫[(2x+1)^4+3x/(x^2+4)]dx =∫(2x+1)^4 dx+∫3x/(x^2+4) dx (u=2x+1 → du=2dx ) =(1/2)∫u^4 du+∫3x/(x^2+4) dx =u^5/10+∫3x/(x^2+4) dx =(2x+1)^5/10+3∫x/(x^2+4) dx (u=x^2+4 → du=2xdx ) =(2x+1)^5/10+3∫(1/u)xdx =(2x+1)^5/10+(3/2)∫(1/u)du =(2x+1)^5/10+(3/2)ln(u)+C =(2x+1)^5/10+(3/2)ln(x^2+4)+C 唔明再問 2012-08-31 00:36:21 補充： 用部分積分法 u和v為函數 ∫udv=uv-∫vdu
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Indefinite integrals Qs
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http://desmond.imageshack.us/Himg32/scaled.php?server=32&filename=96457156.png&res=landing i don't know how to do the above qs i know(1a+1b ) , but i don't know how to use a to do b thank you for ur help ：Ｄ 更新: Ps. The correct website : http://desmond.imageshack.us/Himg444/scaled.php?server=444&filename=87943396.png&res=landing 更新 2: THANKS!最佳解答:
1a. (d/dx)(x^5lnx) { (g×k)'=g'×k+g×k' } =lnx?(d/dx)x^5 + x^5?(d/dx)lnx =5x^4?lnx+x^5?(1/x) =5x^4?lnx+x^4 1b. ∫x^4lnxdx =∫lnxd(x^5/5) =(1/5)∫lnxdx^5 =(1/5){x^5lnx-∫x^5dlnx } =(1/5){x^5lnx-∫x^5/xdx } =(1/5){x^5lnx-∫x^4dx } =(1/5){x^5lnx-x^5/5}+C =(1/25)(5x^5lnx-x^5)+C 2a. (d/dx)(2∫x^4 dx) =(d/dx)[(2/5)x^5+C] =(10/5)x^4 =2x^4 2b. ∫xlnxdx =∫lnxd(x^2/2) =(1/2)∫lnxdx^2 =(1/2)[x^2lnx-∫x^2dlnx] =(1/2)[x^2lnx-∫xdx] =(1/2)[x^2lnx-x^2/2]+C =(1/4)(2x^2lnx-x^2)+C 3. ∫(lnx+1)/2x dx =(1/2)∫(lnx+1)/x dx =(1/2)∫(lnx+1)dlnx =(1/2)[(lnx)^2/2+lnx)+C =(lnx/4)(lnx+2)+C 4. ∫x[(2x+7)^(1/2)-x] dx ( u=2x+7 → x=(u-7)/2 →du=2dx =(1/2)∫[(u-7)/2][√(u)-(u-7)/2] 2dx =(1/2)∫[(u-7)/2][√(u)-(u-7)/2]du =(1/4)∫[u^(3/2)-7u^(1/2)-(u^2-14u+49)/2]du =(1/8)∫[2u^(3/2)-14u^(1/2)-(u^2-14u+49)]du =(1/8)∫[2u^(3/2)-14u^(1/2)-u^2+14u-49]du =(1/8)[(4/5)u^(5/2)-(28/3)u^(3/2)-u^3/3+7u^2-49u]+C 自己代u= = 5. ∫[(2x+1)^4+3x/(x^2+4)]dx =∫(2x+1)^4 dx+∫3x/(x^2+4) dx (u=2x+1 → du=2dx ) =(1/2)∫u^4 du+∫3x/(x^2+4) dx =u^5/10+∫3x/(x^2+4) dx =(2x+1)^5/10+3∫x/(x^2+4) dx (u=x^2+4 → du=2xdx ) =(2x+1)^5/10+3∫(1/u)xdx =(2x+1)^5/10+(3/2)∫(1/u)du =(2x+1)^5/10+(3/2)ln(u)+C =(2x+1)^5/10+(3/2)ln(x^2+4)+C 唔明再問 2012-08-31 00:36:21 補充： 用部分積分法 u和v為函數 ∫udv=uv-∫vdu
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