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

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

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integral of 1/(n(ln(n))^2)

\int \frac{1}{n ( ln ( n ) ) ^{2}}

\frac{x}{n ln ^{2} ( n )} + C

Solution steps

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

Integral of a constant:

\int a d x = a x

= \frac{1}{n ( ln ( n ) ) ^{2}} x

Simplify

\frac{1}{n ln ^{2} ( n )} x : \frac{x}{n ln ^{2} ( n )}

= \frac{x}{n ln ^{2} ( n )}

Add a constant to the solution

= \frac{x}{n ln ^{2} ( n )} + C

\int_{0}^{1} (3 - 4 x) e^{2 x} d x

y^{^{''}} + \frac{200}{8} \cdot y = 30 sin (2 x), y (0) = 2, y^{^{'}} (0) = 0

\frac{d}{d x} (- \frac{c}{x ^{2}})

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

3 t^{4} e^{- 2 t}

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