Integral of 6/(8x-1)

\int\frac{6}{8x-1}dx

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Answer

$\frac{-\frac{3}{4}}{8x-1}+C_0$

Step by step solution

Problem

$\int\frac{6}{8x-1}dx$
1

Solve the integral $\int\frac{6}{8x-1}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=8x-1 \\ du=8dx\end{matrix}$
2

Isolate $dx$ in the previous equation

$\frac{du}{8}=dx$
3

Substituting $u$ and $dx$ in the integral

$\int\frac{\frac{3}{4}}{u^2}du$
4

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\int\frac{3}{4}u^{-2}du$
5

Taking the constant out of the integral

$\frac{3}{4}\int u^{-2}du$
6

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\frac{3}{4}\cdot\frac{u^{-1}}{-1}$
7

Substitute $u$ back for it's value, $8x-1$

$\frac{3}{4}\cdot\frac{\left(8x-1\right)^{-1}}{-1}$
8

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{3}{4}\cdot\frac{\frac{1}{8x-1}}{-1}$
9

Simplify the fraction

$-\frac{3}{4}\cdot\frac{1}{8x-1}$
10

Apply the formula: $a\frac{1}{x}$$=\frac{a}{x}$, where $a=-\frac{3}{4}$ and $x=8x-1$

$\frac{-\frac{3}{4}}{8x-1}$
11

Add the constant of integration

$\frac{-\frac{3}{4}}{8x-1}+C_0$

Answer

$\frac{-\frac{3}{4}}{8x-1}+C_0$

Problem Analysis

Main topic:

Integration by substitution

Time to solve it:

0.35 seconds

Views:

105