INtegration by part

[full glen]

I don't understand last part ı draw it

[full glen]

ı dont understand the last part

how u.arcsinu - ........

queal to

root(1-lnx) + ......

@quick spire

[quick spire]

they do a substitution $t = 1-u^2$ so $dt = -2u du$ and the integral becomes
$$ \int \frac{u}{\sqrt{1-u^2}} du = -\frac{1}{2} \int \frac{1}{\sqrt{t}}dt$$

[proper abyss]

zlik

[quick spire]

Dos this help?

[full glen]

ı am thinking

let me give 2 minutes

🙂

[quick spire]

for $\int \arcsin(u) du$ they use integration by parts. Firstly, they write the integral as $$\int 1\cdot \arcsin(u)du$$ Then they take the derivative of $\arcsin(u)$ which is $\frac{1}{\sqrt{1-u^2}}$ and the anti-derivative of $1$ which is $u$. Therefore
$$\int \arcsin(u) du$$ $$= \int 1\cdot \arcsin(u)du$$ $$= u\cdot \arcsin(u) - \int \frac{u}{\sqrt{1-u^2}} du$$

[proper abyss]

zlik

[quick spire]

Do you understand it?

@full glen

[full glen]

ohhhhh

ı am so sory

yes ı understand

ı forget the write