Integral of (2x)/(x^2^0.5-3)

\int\frac{2x}{\sqrt{x^2}-3}dx

Go!
1
2
3
4
5
6
7
8
9
0
x
y
(◻)
◻/◻
2

e
π
ln
log
lim
d/dx
d/dx
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

Answer

$6\ln\left|x-3\right|-6+2x+C_0$

Step by step solution

Problem

$\int\frac{2x}{\sqrt{x^2}-3}dx$
1

Applying the power of a power property

$\int\frac{2x}{x-3}dx$
2

Taking the constant out of the integral

$2\int\frac{x}{x-3}dx$
3

Solve the integral $\int\frac{x}{x-3}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=x-3 \\ du=dx\end{matrix}$
4

Rewriting $x$ in terms of $u$

$x=3+u$
5

Substituting $u$, $dx$ and $x$ in the integral

$2\int\frac{3+u}{u}du$
6

Split the fraction $\frac{u+3}{u}$ in two terms with same denominator

$2\int\left(\frac{3}{u}+\frac{u}{u}\right)du$
7

Simplifying the fraction by $u$

$2\int\left(\frac{3}{u}+1\right)du$
8

The integral of a sum of two or more functions is equal to the sum of their integrals

$2\left(\int\frac{3}{u}du+\int1du\right)$
9

The integral of a constant is equal to the constant times the integral's variable

$2\left(\int\frac{3}{u}du+u\right)$
10

Substitute $u$ back for it's value, $x-3$

$2\left(\int\frac{3}{u}du+x-3\right)$
11

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

$2\left(3\ln\left|u\right|-3+x\right)$
12

Substitute $u$ back for it's value, $x-3$

$2\left(3\ln\left|x-3\right|-3+x\right)$
13

Multiply $\left(x+-3\right)$ by $2$

$6\ln\left|x-3\right|-6+2x$
14

Add the constant of integration

$6\ln\left|x-3\right|-6+2x+C_0$

Answer

$6\ln\left|x-3\right|-6+2x+C_0$

Problem Analysis

Main topic:

Integration by substitution

Time to solve it:

0.44 seconds

Views:

119