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

The integral of 1/(x(1+ln^2(x))) is arctan(ln(x))+C

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integral of 1/(x(1+ln^2(x)))

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

arctan (ln (x)) + C

Solution steps

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

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 = arctan (u)

= arctan (u)

Substitute back

u = ln (x)

= arctan (ln (x))

Add a constant to the solution

= arctan (ln (x)) + C

\int_{2}^{6} π (y - 2) 2 d y

f (x) = 5 x, at x = 16

y^{^{''}} + 2 y^{^{'}} + y = 2 e^{- t}

\frac{x + 2}{x - 1}

\frac{d y}{d x} = \frac{cos ( 5 x )}{e ^{5 y}}

What is the integral of 1/(x(1+ln^2(x))) ?
The integral of 1/(x(1+ln^2(x))) is arctan(ln(x))+C