Need some help with the following
1. $\displaystyle \int ln(2x+1) dx$
2. $\displaystyle \int tan^{-1}5t dt$
3. $\displaystyle \int (lnx)^2 dx$
Hello, c_323_h!
If you're doing these "by parts", there isn't any choice.
. . So what the problem?
$\displaystyle 1)\;\;\int \ln(2x+1)\,dx$
Let: $\displaystyle u = \ln(2x+1)\qquad\quad dv = dx$
Then: $\displaystyle du = \frac{2}{2x+1}\,dx\qquad v = x$
And we have: .$\displaystyle x\cdot\ln(2x+1) - \int\frac{2x}{2x+1}\,dx$
. . . $\displaystyle =\;x\cdot\ln(2x + 1) - \int\left(1 - \frac{1}{2x + 1}\right)\,dx$
Can you finish it now?
$\displaystyle 2)\;\;\int \tan^{-1}(5t)\,dt$
Let: $\displaystyle u = \tan^{-1}(5x)\qquad\qquad dv = dx$
Then: $\displaystyle du = \frac{5}{1 + 25x^2}\,dx\qquad v = x$
And we have: .$\displaystyle x\cdot\tan^{-1}(5x) - \int\frac{5x}{1 + 25x^2}\,dx\quad\hdots\;etc.$
$\displaystyle 3)\;\;\int (\ln x)^2\,dx$
Let: $\displaystyle u = (\ln x)^2\qquad\qquad dv = dx$
Then: $\displaystyle du = \frac{2\ln x}{x}\,dx\qquad v = x$
And we have: .$\displaystyle x\cdot(\ln x)^2 - \int x\cdot\frac{2\ln x}{x}\,dx$
. . . $\displaystyle = \;x\cdot(\ln x)^2 - 2\underbrace{\int\ln x\,dx}$
. . . . . . . . . . . . . and this can be done by parts