∫√x2+1dxxđặt √x2+1=u⇒x2=u2−1 và 2xdx=2udu$\int\limits\frac{u^{2}du}{u^{2}-1}=\int\limits(1+\frac{u+1-(u-1)}{2(u-1)(u+1)})du=u+\frac{1}{2}\int\limits(\frac{1}{u-1}+\frac{1}{u+1})du$$=u +\frac{1}{2}(ln|u-1|+ln|u+1|)$oi bạn tự thay nhá !!!
∫√x2+1dxxđặt
√x2+1=u⇒x2=u2−1 và
2xdx=2udu$\int\limits\frac{u^{2}du}{u^{2}-1}=\int\limits(1+\frac{u+1-(u-1)}{2(u-1)(u+1)})du=u+\frac{1}{2}\int\limits(\frac{1}{u-1}
-\frac{1}{u+1})du$$=u +\frac{1}{2}(ln|u-1|
-ln|u+1|)
+C $oi bạn tự thay nhá !!!