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
∫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)...
Based on the Integration by parts,:{eq}\displaystyle \int udv = uv-\int vdu... See full answer below.Become a member and unlock all Study Answers Start today. Try it now Create an account Ask a question Our experts can answer your tough homework and study questions. Ask a question...
Letting {eq}u = f(x), v = g(x) {/eq} results in the more common form of the integration by parts formula: {eq}\displaystyle\int udv = uv - \int vdu {/eq} Answer and Explanation: Become a Study.com member to unlock this answer! Create your account View this answer We wi...
∫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...