# Homework Help: Trigonometric substitution

1. Mar 23, 2010

### nameVoid

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