Question

Evaluate the integral with Trig substitution:
integral of x/sqrt((x^2)+x+1) dx

Answer 1

\(\int \dfrac{x}{\sqrt{x^2+x+1}} dx\) ??

thought about completing the square in the denominator?

\(\int \dfrac{x}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}} dx\) ??

then something like u = x + 1/2 ??

Answer 2

Yeah I've did that, the part I get stuck on is the part after that

Answer 3

I've simplified it to say \[\int\limits (2u-1)/\sqrt (4u^2+3)\]

Answer 4

well, if \(\int \dfrac{x}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}} dx\) works, i'd go \( u = x + \frac{1}{2}, \quad du = dx\)

that leads to

\(\int \dfrac{u - \frac{1}{2}}{\sqrt{u^2+\frac{3}{4}}} du\)

\(= \int \dfrac{u}{\sqrt{u^2+\frac{3}{4}}} du - \int \dfrac{\frac{1}{2}}{\sqrt{u^2+\frac{3}{4}}} du\)

**if** the algebra is correct, that is now only a few lines away from completion :p

**if** !!

Answer 5

okay yeah, I got the correct answer, thank you!