∫x+1dx

印轻超1232∫1/x+1dx=?大概是这样 -
贝唯哪19577581093 ______ ∫1/x+1dx=?大概是这样 lnlx+1l+C

印轻超1232∫x/x+1dx怎么化 -
贝唯哪19577581093 ______ x /(x+1) = (x+1-1)/(x+1) =1 - 1/(x+1) I = x - ∫dx(x+1) =x - ln|x+1| +C

印轻超1232计算不定积分∫x√x2+1dx -
贝唯哪19577581093 ______[答案] ∫x√(x^2+1)dx =(1/2)∫√(x^2+1)dx^2 =(1/2)∫(x^2+1)^(1/2)d(x^2+1) =(1/2)*[(x^2+1)^(1/2+1)/(1/2+1)]+C =(1/3)(x^2+1)^(3/2)+C =(1/3)*(x^2+1)/√(x^2+1)+C

印轻超1232∫√x+1/xdx -
贝唯哪19577581093 ______ 设√x+1=t x=t²-1 ∫√x+1/xdx=∫t/(t²-1)d(t²-1) =2∫t²/(t²-1)dt =2∫(t²-1+1)/(t²-1)dt =2[∫1dt+∫1/(t²-1)dt] =2t+∫(t+1-t+1)/(t+1)(t-1)dt =2t+ln(t-1)-ln(t+1)

印轻超1232∫(xcosx+1)dx 上限是1 下限是 - 1 -
贝唯哪19577581093 ______ ∫(-1,1)(xcosx+1)dx =∫ xcosx dx + ∫1dx =∫ xdsinx + x|(-1,1) =xsinx|(-1,1) - ∫sinxdx +2 =sin1 + sin-1 + 2 +cosx|(-1,1) =sin1 + sin-1 + cos1 - cos-1 + 2

印轻超1232∫ 10xx+1dx=______. -
贝唯哪19577581093 ______[答案] ∫ 10 x x+1dx=∫ 10(1- 1 x+1)dx=[x-ln(x+1)] |10=1-ln2 故答案为:1+ln2

印轻超1232∫x√x+1dx (x根号x+1 dx)求不定积分. -
贝唯哪19577581093 ______[答案] 令√(x+1)=u,则x=u²-1,dx=2udu 原式=∫ (u²-1)*u*2udu =2∫ (u^4-u²)du =(2/5)u^5-(2/3)u³+C =(2/5)(x+1)^(5/2)-(2/3)(x+1)^(3/2)+C

印轻超1232求不定积分∫cos根号下x+1dx -
贝唯哪19577581093 ______[答案] ∫cos√(x+1)d(x+1) =∫cos√(x+1)d(√(x+1)^2 √(x+1)=t =∫costdt^2=∫2tcostdt=∫2tdsint=2tsint-2∫sintdt=2tsint+2cost+C =2√(x+1)sin√(x+1)+2cos√(x+1) +C

印轻超1232计算∫上1下0,x根号下x+1dx -
贝唯哪19577581093 ______[答案] =∫x(根号下x+1)dx =∫(x+1)(根号下x+1)d(x +1)-∫(根号下x+1)d(x +1)=2/5*(x +1)^(5/2)-2/3*(x +1)^(3/2)=4(√2+1)/15

印轻超1232∫1/x∧4+1dx -
贝唯哪19577581093 ______ sec²x=1+tan²x 设x=sec²t 代入看看