Question

数学不定积分，需要用分部积分法解

求 (根号X) * arctan (根号X）dx的不定积分

Answer 1

求不定积分∫(√x)arctan(√x)dx
解：令arctan(√x)=u，则√x=tanu，x=tan²u，dx=2tanusec²udu；
故原式=2∫utan²usec²udu=(2/3)∫ud(tan³u)=(2/3)[utan³u-∫tan³udu]
=(2/3)[utan³u-∫tanu(1-sec²u)du]=(2/3)[utan³u-∫tanudu+∫tanusec²udu]
=(2/3)[utan³u+∫d(cosu)/cosu+∫tanud(tanu)]
=(2/3)[utan³u+ln∣cosu∣+(1/2)tan²u]+C
=(2/3){x³/²arctan(√x)+ln[1/√(1+x)]+x/2}+C
=(2/3)[(x√x)arctan(√x)-(1/2)ln(1+x)+x/2]+C

Answer 2

设 x = (tant)^2，则 dx = 2tant*(sect)^2 *dt
因此，原积分变换为：
=∫(tant) * t * dx
=∫(tant) * t * 2tant * (sect)^2 * dt
=2∫t * (tant)^2 * (sect)^2 * dt
为了计算方便，再设 u = t，dv = (tant)^2 *(sect)^2 *dt，则 du = dt，v = ∫(tant)^2 * d(tant) =1/3(tant)^3。使用分部积分，得到：
=2∫u*dv
=2u*v - 2∫v*du
=2t * [1/3 * (tant)^3] - 2/3*∫(tant)^3 *dt
=2/3 * t* (tant)^3 - 2/3 * ∫tant * (tant)^2 * dt
=2/3 *t * (tant)^3 - 2/3 * ∫tant * [(sect)^2 - 1] *dt
=2/3 *t * (tant)^3 - 2/3 * ∫tant * d(tant) + 2/3 *∫tant * dt
=2/3 *t * (tant)^3 - 1/3 * (tant)^2 + 2/3 * ∫ -d(cost)
=2/3 *t * (tant)^3 - 1/3 * (tant)^2 - 2/3 *ln|cost| + C
=2/3* arctan√x * (√x)^3 - x/3 + 1/3 *ln(sect)^2 + C 注：-2/3*ln|cost| = 1/3*ln|1/(cost)^2| = 1/3ln(sect)^2
=2/3*arctan√x * (√x)^3 - x/3 + 1/3 *ln(x + 1) + C

Answer 3

∫√x*arctan(√x) dx = 2/3 * x^(3/2) * arctan(√x) - ∫2/3 * x^(3/2) (arctan √x)' dx
(arctan √x)' = (√x)'/1+x=1/2 * x^(-1/2)/(1+x)
所以 ∫√x*arctan(√x) dx = 2/3 * x^(3/2) * arctan(√x) - ∫1/3 * x/(1+x) = 2/3 * x^(3/2) * arctan(√x) - 1/3[x-ln(x+1)] +c

Answer 4