f(x)=1/((5x-1)ln(5x-1))−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

1(5x1)(ln(5x1))dx\int\frac{1}{\left(5x-1\right)\left(\ln\left(5x-1\right)\right)}dx

Step-by-step Solution

1

We can solve the integral 1(5x1)ln(5x1)dx\int\frac{1}{\left(5x-1\right)\ln\left(5x-1\right)}dx by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it uu), which when substituted makes the integral easier. We see that 5x15x-1 it's a good candidate for substitution. Let's define a variable uu and assign it to the choosen part

u=5x1u=5x-1
2

Now, in order to rewrite dxdx in terms of dudu, we need to find the derivative of uu. We need to calculate dudu, we can do that by finding the derivative of the equation above

du=5dxdu=5dx
3

Isolate dxdx in the previous equation

dx=du5dx=\frac{du}{5}
4

Substituting uu and dxdx in the integral and simplify

151uln(u)du\frac{1}{5}\int\frac{1}{u\ln\left(u\right)}du
5

We can solve the integral 1uln(u)du\int\frac{1}{u\ln\left(u\right)}du by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it vv), which when substituted makes the integral easier. We see that ln(u)\ln\left(u\right) it's a good candidate for substitution. Let's define a variable vv and assign it to the choosen part

v=ln(u)v=\ln\left(u\right)
6

Now, in order to rewrite dudu in terms of dvdv, we need to find the derivative of vv. We need to calculate dvdv, we can do that by finding the derivative of the equation above

dv=1ududv=\frac{1}{u}du
7

Isolate dudu in the previous equation

du=udvdu=udv
8

Substituting vv and dudu in the integral and simplify

151vdv\frac{1}{5}\int\frac{1}{v}dv
9

The integral of the inverse of the lineal function is given by the following formula, 1xdx=ln(x)\displaystyle\int\frac{1}{x}dx=\ln(x)

15lnv\frac{1}{5}\ln\left|v\right|
10

Replace vv with the value that we assigned to it in the beginning: ln(u)\ln\left(u\right)

15lnlnu\frac{1}{5}\ln\left|\ln\left|u\right|\right|
11

Replace uu with the value that we assigned to it in the beginning: 5x15x-1

15lnln5x1\frac{1}{5}\ln\left|\ln\left|5x-1\right|\right|
12

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration CC

15lnln5x1+C0\frac{1}{5}\ln\left|\ln\left|5x-1\right|\right|+C_0

Final answer to the exercise

15lnln5x1+C0\frac{1}{5}\ln\left|\ln\left|5x-1\right|\right|+C_0

Try other ways to solve this exercise

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