Question

the definite integral of ln(1+t^2) from (sqrt(1-x) to 0)

Answer 1

\[\large \int\limits_{\sqrt{1-x}}^{0}\ln(1+t^2)dt\]integration by parts. \[\large{u= \ln(1+t^2)\quad\quad\quad dv= dt \\du =\frac{2t}{1+ t^2}dt\quad\quad\quad v=t}\]\[\large t \ln(1+t^2)-\int\limits t*(\frac{2t}{1+t^2})dt\]evaluate integral first. \[\large -\int\limits \frac{2t^2}{1+t^2}dt =- 2 \int\limits \frac{t^2}{1+t^2}dt\]Use long division to break down the stuff inside the integral. As shown here. http://www.sketchtoy.com/37477197 \[\large -2 \int\limits (1-\frac{1}{1+t^2})dt\]\[\large -2 \ [\int\limits\limits dt - \int\limits\limits (\frac{1}{1+t^2})dt ]=-2t +2\tan^{-1}(t) \] Combine it with the other part \[\large t \ln(1+t^2)-[-2t+2\tan^{-1}(t)]\]\[\huge t \ln(1+t^2)+2t-2\tan^{-1}(t)]_{\sqrt{1-x}}^{0}\]\[\large 0 - [\sqrt{1-x}* \ln(2-x)+2\sqrt{1-x}-2\tan^{-1}(\sqrt{1-x})]\]\[\large -\sqrt{1-x}* \ln(2-x-2\sqrt{1-x}+2\tan^{-1}(\sqrt{1-x})\]

Answer 2

@Jhannybean avoid using \(*\) in place of \(\cdot\), \(\times\), or just implicitly signifying multiplication like $$-\sqrt{1-x}\log\left(2-x-2\sqrt{1-x}+2\arctan\sqrt{1-x}\right)$$

Answer 3

\(*\) is used to denote an operation called convolution often.

Answer 4

idk howto do the dot!!!! :(

Answer 5

It's so frustrating ~.~

Answer 6

$$\cdot=\text{\cdot}$$

Answer 7

Thanks :D haha. That made my life a whole lot easier....\[ \bf \cdots \ \text{winning!!} \cdots\]