Question

Integral 1/x(log x)²

Answer 1

Solution:

Given Integral:

$\displaystyle \tt \longrightarrow I = \int \frac{dx}{x \: ln^{2}(x)}$

We will solve this problem by substitution method.

Let us assume that:

$\tt \longrightarrow u =ln(x)$

$\tt \longrightarrow du = \dfrac{dx}{x}$

So, the integral changes to:

$\displaystyle \tt \longrightarrow I = \int \frac{du}{u^{2}}$

Now, we know that:

$\displaystyle \tt \longrightarrow\int {u}^{n} \: du = \dfrac{ {u}^{n + 1} }{n + 1} + C$

So, we get:

$\displaystyle \tt \longrightarrow I = \dfrac{ - 1}{u} + C$

Substituting back u = ln(x), we get:

$\displaystyle \tt \longrightarrow I = \dfrac{ - 1}{ln(x)} + C$

Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$

Answer 2

Explanation:

Given Integral:

\displaystyle \tt \longrightarrow I = \int \frac{dx}{x \: ln^{2}(x)}⟶I=∫

xln

2

(x)

dx

We will solve this problem by substitution method.

Let us assume that:

\tt \longrightarrow u =ln(x)⟶u=ln(x)

\tt \longrightarrow du = \dfrac{dx}{x}⟶du=

x

dx

So, the integral changes to:

\displaystyle \tt \longrightarrow I = \int \frac{du}{u^{2}}⟶I=∫

u

2

du

Now, we know that:

\displaystyle \tt \longrightarrow\int {u}^{n} \: du = \dfrac{ {u}^{n + 1} }{n + 1} + C⟶∫u

n

du=

n+1

u

n+1

+C

So, we get:

\displaystyle \tt \longrightarrow I = \dfrac{ - 1}{u} + C⟶I=

u

−1

+C

Substituting back u = ln(x), we get:

\displaystyle \tt \longrightarrow I = \dfrac{ - 1}{ln(x)} + C⟶I=

ln(x)

−1

+C

Which is our required answer.

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