$\displaystyle \int \left( \frac{1}{x^2+3} \right)\; dx$
I've let $u=x^2+3$ but can't seem to get the right answer.
Really not sure what to do.
$\displaystyle \int \left( \frac{1}{x^2+3} \right)\; dx$
I've let $u=x^2+3$ but can't seem to get the right answer.
Really not sure what to do.
Hint: Note that $\frac{1}{x^{2}+3}=\frac{1}{3}\cdot \frac{1}{\left( \frac{x}{\sqrt{% 3}}\right) ^{2}+1}$
and use the substitution $u=\frac{x}{\sqrt{3}}$.
Or please try putting $x = \sqrt{3} \tan\theta$. Then you have $dx = \sqrt{3}\cdot\sec^{2}\theta \ d\theta$. So you have your integral as:
\begin{align*} \int\frac{1}{x^{2}+3} \ \text{dx} &= \int \frac{\sqrt{3} \cdot \sec^{2}\theta}{3 (1+\tan^{2}\theta)} \ \text{d}\theta \\ &=\frac{1}{\sqrt{3}} \int \text{d}\theta \end{align*}
I suggest letting $u = \frac{x}{\sqrt{3}}$. Then $du = \frac 1{\sqrt{3}} dx$ and $ \int \left( \frac 1{x^2+3} \right) dx = \int \left( \frac 1{(\sqrt 3 u)^2 + 3} \, \sqrt 3 \right) dx = \frac 1{\sqrt{3}} \int \left( \frac 1{u^2 + 1} \right) = \frac 1{\sqrt{3}} \arctan(u) + C. $ Hope that helps