Trigonometric substitution

[nameVoid]

$$\int\frac{x}{\sqrt{x^2+x+1}}dx$$
$$\int \frac{x}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}}dx$$
$$u=x+\frac{1}{2}$$
$$\int \frac{u-\frac{1}{2}}{\sqrt{u^2+\frac{3}{4}}}du$$
$$u=\frac{\sqrt{3}}{2}tanT$$
$$du=\frac{\sqrt{3}}{2}sec^2TdT$$
$$\int \frac{\frac{(\sqrt{3}}{2}tanT-\frac{1}{2})\frac{\sqrt{3}}{2}sec^2T}{\frac{\sqrt{3}}{2}secT}dT$$
$$\int (\frac{\sqrt{3}}{2}tanT-\frac{1}{2})secTdT$$
$$\frac{\sqrt{3}}{2}secT-\frac{1}{2}ln|secT+tanT|+C$$
$$\frac{\sqrt{u^2+\frac{3}{4}}}{2}-\frac{ln|\sqrt{u^2+\frac{3}{4}}+2u|}{2}+C$$
$$\frac{\sqrt{x^2+x+1}}{2}-\frac{ln|\sqrt{x^2+x+1}-2x+1|}{2}$$
​

 

[LudusRex]

+C


Therefore, the integral can be rewritten as:


\int\frac{x}{\sqrt{x^2+x+1}}dx = \frac{\sqrt{x^2+x+1}}{2}-\frac{ln|\sqrt{x^2+x+1}-2x+1|}{2}+C


Trigonometric substitution is a powerful technique used in calculus to solve integrals that involve expressions containing trigonometric functions. In this case, we used the substitution u=x+1/2 to simplify the integral and make it easier to solve using trigonometric identities. By applying the appropriate substitutions and using fundamental trigonometric identities, we were able to solve the integral and provide an exact expression for the solution. This method is often used in more complex integrals and can be a useful tool for solving challenging mathematical problems.