Which of the following represents the integral ∫x1+x2dx\displaystyle{\int \frac{x}{\sqrt{1+x^2}}dx}∫1+x2xdx
1) x21+x+C\dfrac{x^2}{\sqrt{1+x}}+C1+xx2+C
2) ln(1+x2)+C\ln{(1+x^2)}+Cln(1+x2)+C
3) 1+x2+C\sqrt{1+x^2}+C1+x2+C
4) x21+x2+C\dfrac{x^2}{\sqrt{1+x^2}}+C1+x2x2+C
We rewrite the integral as: ∫x1+x2dx=12∫d(1+x2)1+x2\int\frac{x}{\sqrt{1+x^2}}dx=\frac{1}{2}\int\frac{d(1+x^2)}{\sqrt{1+x^2}}∫1+x2xdx=21∫1+x2d(1+x2) and make the change of variables: z=1+x2z=1+x^2z=1+x2. We receive: ∫x1+x2dx=12∫d(1+x2)1+x2=12∫dzz\int\frac{x}{\sqrt{1+x^2}}dx=\frac{1}{2}\int\frac{d(1+x^2)}{\sqrt{1+x^2}}=\frac{1}{2}\int\frac{dz}{\sqrt{z}}∫1+x2xdx=21∫1+x2d(1+x2)=21∫zdz
We integrate the latter and receive: 12∫dzz=z+C=1+x2+C,C∈R\frac{1}{2}\int\frac{dz}{\sqrt{z}}=\sqrt{z}+C=\sqrt{1+x^2}+C,C\in{\mathbb{R}}21∫zdz=z+C=1+x2+C,C∈R.
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