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

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

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

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

arctan (ln (x)) + C

Solution steps

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

Simplify

= \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

x^{3}

\int \frac{1}{- x ^{2} + 4 x - 5} d x

\int x cot^{2} (x) d x

\int \frac{x e ^{x}}{e ^{x} - 2} d x

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