For test tomorrow- Integration by Parts

Hello, c_323_h!

If you're doing these "by parts", there isn't any choice.
. . So what the problem?

Quote:

$1)\;\;\int \ln(2x+1)\,dx$

Let: $u = \ln(2x+1)\qquad\quad dv = dx$
Then: $du = \frac{2}{2x+1}\,dx\qquad v = x$

And we have: . $x\cdot\ln(2x+1) - \int\frac{2x}{2x+1}\,dx$

. . . $=\;x\cdot\ln(2x + 1) - \int\left(1 - \frac{1}{2x + 1}\right)\,dx$

Can you finish it now?

Quote:

$2)\;\;\int \tan^{-1}(5t)\,dt$

Let: $u = \tan^{-1}(5x)\qquad\qquad dv = dx$

Then: $du = \frac{5}{1 + 25x^2}\,dx\qquad v = x$

And we have: . $x\cdot\tan^{-1}(5x) - \int\frac{5x}{1 + 25x^2}\,dx\quad\hdots\;etc.$

Quote:

$3)\;\;\int (\ln x)^2\,dx$

Let: $u = (\ln x)^2\qquad\qquad dv = dx$

Then: $du = \frac{2\ln x}{x}\,dx\qquad v = x$

And we have: . $x\cdot(\ln x)^2 - \int x\cdot\frac{2\ln x}{x}\,dx$

. . . $= \;x\cdot(\ln x)^2 - 2\underbrace{\int\ln x\,dx}$
. . . . . . . . . . . . . and this can be done by parts