arctanx∧2的不定积分

方泉盲15401/x^2+1的不定积分
家狐滕18796893655 ______ 1/x^2+1的不定积分是:arctan(x) +C原因如下:令x=tan t,t∈(-π/2,π/2),t= arctan xdx=dt/cos^2 t1/(x^2+1)=1/(tan^2 t+1)=cos^2 t所以∫dx/(x^2+1)=∫(dt/cos^2 t )* cos^2t=∫dt=t+C=...

方泉盲1540解不定积分求∫(√arctanx)/1+x²解法第二步可以看懂,但是怎么变成它的3/2次方呀,不该是∫(arctanx)∧1/2d(arctanx) -
家狐滕18796893655 ______[答案] ∫(√arctanx)/(1+x²)dx =∫(√arctanx)darctanx =2/3(arctanx)^(3/2)+C

方泉盲1540求x(arctanx)的不定积分 -
家狐滕18796893655 ______[答案] ∫x(arctanx)dx =(1/2)∫ (arctanx)d(x^2) = (1/2)x^2(arctanx) -(1/2)∫ x^2 (1/(1+x^2) dx = (1/2)x^2(arctanx) - (1/2)∫ dx+ (1/2)∫ 1/(1+x^2) dx = (1/2)x^2(arctanx) - (1/2)x + (1/2) arctanx + C

方泉盲1540求不定积分 ∫( xarctanx)dx= -
家狐滕18796893655 ______[答案] ∫ xarctanx dx = ∫ arctanx d(x²/2) = (x²/2)arctanx - (1/2)∫ x² d(arctanx) = (1/2)x²arctanx - (1/2)∫ x²/(x² + 1) dx = (1/2)x²arctanx - (1/2)∫ [(x² + 1) - 1]/(x² + 1) dx = (1/2)x²arctanx - (1/2)∫ dx + (1/2)∫ dx/(x² + 1) = (1/2)x²arctanx - x/2 + (1/2)arctanx + C

方泉盲1540求(arctanx)/(x^2*(1+x^2))的不定积分 -
家狐滕18796893655 ______[答案] ∫ tan⁻¹x/[x²(1 + x²)] dx = ∫ tan⁻¹x d(- 1/x - tan⁻¹x) = tan⁻¹x · (- 1/x - tan⁻¹x) - ∫ (- 1/x - tan⁻¹x) d(tan⁻¹x) = - (tan⁻¹x)/x - (tan⁻¹x)² + ∫ (1/x + tan⁻¹x)/(1 + x²) dx = - (tan⁻¹x)/x - (tan⁻¹x)² + ∫ [(1 + x²) - x²]/[x(1 + x²)] + ∫ tan...

方泉盲1540求arctanx/(x2(1+x2))的不定积分? -
家狐滕18796893655 ______[答案] ∫arctanxdx/[x^2(1+x^2)]=∫arctanxdx/x^2 -∫arctanxdx/(1+x^2)=∫arctanxd(-1/x)-∫arctanxdarctanx=-(arctanx)/x +∫(1/x)darctanx-(arctanx)^2/2=-(arctanx)/x-(arctanx)^2/2+∫dx/[x(1+x^2)]其中 ∫dx/[x(1+x^...

方泉盲1540x*arctanx的不定积分是多少 -
家狐滕18796893655 ______[答案] ∫ xarctanx dx = ∫ arctanx d(x²/2) = (x²/2)arctanx - (1/2)∫ x² d(arctanx) = (1/2)x²arctanx - (1/2)∫ x²/(x² + 1) dx = (1/2)x²arctanx - (1/2)∫ [(x² + 1) - 1]/(x² + 1) dx = (1/2)x²arctanx - (1/2)∫ dx + (1/2)∫ dx/(x² + 1) = (1/2)x²arctanx - x/2 + (1/2)arctanx + C

方泉盲1540arctanx的不定积分是什么要快啊, -
家狐滕18796893655 ______[答案] ∫arctanxdx = xarctanx - ∫x d(arctanx) = xarctanx - ∫ x/(1+x²)dx = xarctanx - (1/2)∫1/(1+x²) d(1+x²) = xarctanx - (1/2)ln(1+x²) + C

方泉盲1540arctanx/x^2的不定积分 -
家狐滕18796893655 ______[答案] 用分部积分,设u=arctanx,v'=1/x^2u'=1/(1+x^2),v=-1/x,原式=-(arctanx)/x+∫ dx/[x(1+x^2)]=-(arctanx)/x+∫(-x) dx/(1+x^2)+∫ dx/x=-(arctanx)/x-(1/2)∫d(1+x^2)/(1+x^2)+∫ dx/x=-(arctanx)/x-(1/2)ln(1+x^2)+ln...