I=(1/2)*x^2*ln(x^2+1)-(1/2)*∫x^2*2x*dx/(x^2+1)
=(1/2)*x^2*ln(x^2+1)-∫x^3*dx/(x^2+1)
=(1/2)*x^2*ln(x^2+1)-(1/2)∫(x^2)d(x^2)/(x^2+1)
=(1/2)*x^2*ln(x^2+1)-(1/2)∫[1-1/(x^2+1)]d(x^2)
=(1/2)*[x^2*ln(x^2+1)-x^2+ln(x^2+1)]+c
=(1/2)*[(x^2+1)*ln(x^2+1)-x^2]+c
知道分部积分,为什么不用分部积分的方法做呢,
追问对不起 没看懂
追答![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/7acb0a46f21fbe090e7a25a166600c338744ad5b?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/8718367adab44aed01765851be1c8701a18bfb4f?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
方法步骤已经非常详尽,还临时编了一个,说明分部积分的教材,
因为复制粘贴,不知道哪里会粘贴错吧,
第二张图片的第三行,注意后面的 x 的指数是3,
追问谢谢您 我就差一步化简就和答案一样了