can someone please help?
I need to evaluate this indefinite integral:
$$\int \frac{x}{\sqrt{x^2+2}}dx$$ I tried using substitution for letting u = x, but I can't get past finding the antiderivative after that.
Thank you!
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I need to evaluate this indefinite integral:
$$\int \frac{x}{\sqrt{x^2+2}}dx$$ I tried using substitution for letting u = x, but I can't get past finding the antiderivative after that.
Thank you!
For this you want to set $u = x^2+2$ which means $du=2x*dx$. We see an $x$ in the integrand which is good but we don't see a $2$. We can make up for this by saying.
$$\int{\frac{x}{\sqrt{x^2+2}}} = \frac{1}{2}\int{\frac{2x}{\sqrt{x^2+2}}}$$
And now we can use our $u$ substitution to get.
$$\int{\frac{x}{\sqrt{x^2+2}}}=\frac{1}{2}\int{\frac{1}{\sqrt{u}}du}=\frac{1}{2}\frac{\sqrt{u}}{\frac{1}{2}}+C=\sqrt{u}+C$$
Substituting $x^2+2$ back in for $u$ we find that.
$$\int{\frac{x}{\sqrt{x^2+2}}}=\sqrt{x^2+2}+C$$
Letting $u=x$ doesn't accomplish anything. Try letting $u=x^2+2$.
write the integral inthe form $$\frac{1}{2}\int\frac{2x}{\sqrt{x^2+2}}dx$$ and set $$x^2+2=t$$