Question

求大神解不定积分

10 12 14跪求

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Answer 1

详细过程如图rt所示……希望写的很清楚帮到你

Answer 2

14. (x-2)cos(2x)-sin(2x)/2+C

16. 2sin(sqrt(x))-2sqrt(x)cos(sqrt(x))+C
2. -(x+1)e^(-x)+C

4. (-ln(x)-1)/x+C
6. x*arctan(x)-(ln(x^2+1))/2+C
8. (x^(n+1)*((n+1)*ln(x)-1))/(n+1)^2+C
10. (x-1)*e^x+C
12. (ln((1-x)/(1+x))+x(x*ln((1-x)/(1+x))-2))/2+C

Answer 3

∫e^(x+lnx)dx

=∫(e^x·e^lnx)dx

=∫(xe^x)dx

=∫xd(e^x)

=xe^x-∫e^xdx

=xe^x-e^x+C

∫xln[(1-x)/(1+x)]dx

=∫ln[(1-x)/(1+x)]d(½x²)

=½x²ln[(1-x)/(1+x)]-½∫x²d[ln(1-x)/(1+x)]

=½x²ln[(1-x)/(1+x)]+½∫x²·2/(1-x²)dx

=½x²ln[(1-x)/(1+x)]-∫[(1-x²)-1]/(1-x²) dx

=½x²ln[(1-x)/(1+x)]-∫dx+∫dx/(1-x²)

=½x²ln[(1-x)/(1+x)]-x-½ln|(1-x)/(1+x)|+C

∫[(4-2x)sin2x)]dx

=∫(x-2)d(cos2x)

=∫xd(cos2x)-2cos2x

=xcos2x-∫cos2xdx-2cos2x

=xcos2x-½sin2x-2cos2x+C