What is the integral of 1/(ysqrt(1+(ln(y))^2)) ?

The integral of 1/(ysqrt(1+(ln(y))^2)) is ln|sqrt(ln^2(y)+1)+ln(y)|+C

Popular >

integral of 1/(ysqrt(1+(ln(y))^2))

\int \frac{1}{y 1 + ( ln ( y ) ) ^{2}} d y

ln ln^{2} (y) + 1 + ln (y) + C

Solution steps

\int \frac{1}{y 1 + ( ln ( y ) ) ^{2}} d y

Simplify

= \int \frac{1}{y 1 + ln ^{2} ( y )} d y

Apply u-substitution

: \int \frac{1}{1 + u ^{2}} d u

= \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = ln u^{2} + 1 + u

= ln u^{2} + 1 + u

Substitute back

u = ln (y)

= ln ln^{2} (y) + 1 + ln (y)

Add a constant to the solution

= ln ln^{2} (y) + 1 + ln (y) + C

y = \frac{( x ^{2} + 1 )}{( x ^{3} - 1 )}

\int sec (t) (8 sec (t) + 5 tan (t)) d t

\frac{d}{d x} ((2 x)^{3})

x \to 0 lim ((x)^{3})

\frac{\partial}{\partial x} (3 sin (2 x) cos (6 y))

What is the integral of 1/(ysqrt(1+(ln(y))^2)) ?
The integral of 1/(ysqrt(1+(ln(y))^2)) is ln|sqrt(ln^2(y)+1)+ln(y)|+C