首先根据分部积分法的定义,由以下表达式: \int udv=uv- \int vdu 然后将其表示为: \int \ln xd( \dfrac {1}{2}x^{2}) = \dfrac {1}{2}x^{2} \ln x- \int \dfrac {1}{2}x^{2}d( \ln x) 最后可以得出 u= \ln x,v= \dfrac {x^{2}}{2} 故答案为 u= \ln x,v= \d...
本题为求解不定积分的值,可以用分部积分法 \int udv=uv- \int vdu 求得, \int (2x+3)e^{x}dx= \int 2xe^{x}dx+ \int 3e^{x}dx =2 \int xd(e^{x})+ \int 3e^{x}dx ,进一步化简即可求出本题答案反馈 收藏
Answer and Explanation: Evaluate the given integral. We do this by applying the formula for integration by parts, {eq}\displaystyle \int udv = uv - \int vdu {/eq} where we...Become a member and unlock all Study Answers Start today. Try it now Create an account Ask a question...
Answer and Explanation:1 The formula for Integration by Parts is {eq}\displaystyle \int udv = uv - \int vdu {/eq}. The derivative formulas for hyperbolic cosine and... Learn more about this topic: Hyperbolic Functions: Properties & Applications ...
应用积分公式：接下来，我们应用积分的基本公式，即∫udv = uv ∫vdu，其中u是cos3t，dv是d。由于dv是d，我们可以直接将其积分得到cost，而u的积分为∫cos3tdt的求解则相对复杂，但在这里我们不需要显式地求出它，因为我们可以直接对cost进行积分并乘以cos3t的系数。然而，为了得到最终答案，我们...
16 相关知识点: 试题来源: 解析 ∫udv=uv-∫vdu,在这个公式中,V的值经常是ex,sina,cosa,因为这些在多次分步积分后又周期性还原回来了.至于你所说的,解答省略了一部而已∫2tetdt=∫2tdet=2tet-∫etd(2t)=…… 反馈 收藏
(2)1. -解析 u s ∫_0^12^xe^xdx =∫_0^12^xde^x =2e—ed2分步积分法: ∫udv=uv-∫vdu =2^2e^x|_0^1-e^x_2^xln21^1 =2e(1-1n2)+1n2-1. 2s ∫_(-1)^1|x^2-x|dx x^2-x=x(x-1) 且xc[.1] 得X.0]时x ∈[-1, x^2-x≥0 x[0.1]时x x^2-x≤0 ∴...
∫udv=uv−∫vdu\int_{}^{}udv=uv-\int_{}^{}vdu 5.1I=∫x+cosx1+sinxdx5.1I=\int_{}^{}\frac{x+cosx}{1+sinx}dx I=∫[x(sinx2+cosx2)2+cosx1+sinx]dx=12∫x[sec(x2−π4)]2dx+ln|1+sinx|=∫xd[(tan(x2−π4)]+ln|1+sinx|=xtan(x2−π4)+2ln|cos(x2−π4...
∫udv=uv−∫vdu\int_{}^{}udv=uv-\int_{}^{}vdu 5.1I=∫x+cosx1+sinxdx5.1I=\int_{}^{}\frac{x+cosx}{1+sinx}dx I=∫[x(sinx2+cosx2)2+cosx1+sinx]dx=12∫x[sec(x2−π4)]2dx+ln|1+sinx|=∫xd[(tan(x2−π4)]+ln|1+sinx|=xtan(x2−π4)+2ln|cos(x2−π4...
The solution for the integral by using the integrating by parts rule is obtained by using the below formula, {eq}\int {udv = uv - \int {vdu} } {/eq} If the given integral is of the form of {eq}\int {udv} {/eq} then we can use the above formula to find the solution for ...