∫ √(9+25x^2) dx帮忙解答一下,请带步骤.谢谢~!
1个回答
由于f(x)为偶函数,故可设x>0,(x0积分一样).设x=(3/5)*tanα,则dx=(3/5)*(secα)^2dα 则原积分=9/5*∫√(1+(tanα)^2) * (secα)^2dα =9/5*∫(secα)^3dα 求I=∫(secα)^3dα,I=∫secα dtanα =secα*tanα-∫tanαdsecα =secα*tanα-∫(tanα)^2secαdα =secα*tanα-∫(secα)^3dα+∫secαdα 故I=(secα*tanα+ln|secα+tanα|)/2+C secα=√(1+(tanα)^2)=√(9+25x^2)/3.tanα=5x/3.所以原积分=x*√(9+25x^2)/2+(9/10)*ln(5x/3 +√(9+25x^2)/3)+C =x*√(9+25x^2)/2+(9/10)*arcsinh(5x/3)+C.(注意是arcsinh,不是arcsin).
