1,令t=2^x于是有x=log2(t),所以有dx=dt/(tln2)代入到积分得到∫ 2^x/(1+4^x) dx =∫dt/(1+t^2)ln2=(1/ln2)arctant+c
将t=2^x代入得到∫ 2^x/(1+4^x) dx =(1/ln2)arctan2^x+c;
2,方法类似于1.
∫(X+1)/(x^2+9) dx =∫xdx/(x^2+9)+∫dx/(x^2+9)
=1/2∫d(x^2+9)/(x^2+9)+1/3∫d(x/3)/[1+(x/3)^2
=[ln(x^2+9)]/2+1/3arctan(x/3)+c;
3,令tg√x=t于是可以得到x=(arctant)^2,所以有dx=2arctant/(1+t^2)代入到积分式中得到∫(tg根号X) / 根号X dx
=∫2tdt/(1+t^2)
=∫d(t^2+1)/(t^2+1)
=ln(t^2+1)+c.
将t=tg√x代回到上式中得到∫(tg根号X) / 根号X dx=ln[(tg√x)^2+1]+c;