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...
$$ $$ V_{1}- \frac{1}{2}e^{-x^{2}} $$ $$ \intudv=uv- \int vdu \\ =- \frac{x^{2}}{2}e^{-x^{2}}+ \int xe^{-x^{2}}dx \\ =- \frac{x^{2}}{2}e^{-x^{2}}- \frac{1}{2}e^{-x^{2}}+c $$ $$ \sqrt{1^{\sqrt{12}2}}^...
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 ...
∫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...
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
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
∫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...
One technique of integration is called the integration by parts. This method is typically applied to integrands that are composed of a product of two distinct functions. We apply the formula, {eq}\displaystyle \int udv = uv-\int vdu {/eq}, and we assign {eq}\displays...
The integration technique of integration by parts is quite applicable when we can consider the integrand as a product of two distinct terms. These terms would be assigned to the variablesuanddvin the formula, $$\displaystyle \int udv = uv - \int vdu $$ ...