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

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

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

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

arctan (ln (x)) + C

Solution steps

\int \frac{1}{x + x 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

\frac{d x}{d t} + 4 t^{5} x^{7} + \frac{x}{t} = 0

y^{^{''}} - \frac{2}{x ^{2}} y = 0, y (- 1) = - 1, y^{^{'}} (- 1) = - 4

\int (9 t^{2} + \frac{1}{t ^{3}}) d t

y^{^{'}} = e^{3 x} - y

\int 3 tan^{3} (x) d x

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