Integral Exercise 5
We’re asked to evaluate the integral
$$ \int \frac{1}{x \sqrt{2x-1}} dx $$
To solve it, we’ll use the substitution method.
There are two different substitution paths we can take - both will lead to the same result.
Method 1
The main difficulty here lies in the square root expression:
$$ \int \frac{1}{x \sqrt{2x-1}} dx $$
Let’s define the square root as our new variable:
$$ t = \sqrt{2x-1} $$
Next, we differentiate both sides with respect to their respective variables:
$$ D[t] \ dt = D[\sqrt{2x-1}] \ dx $$
$$ 1 \ dt = \frac{1}{2 \sqrt{2x-1}} \cdot 2 \ dx $$
$$ dt = \frac{1}{\sqrt{2x-1}} \ dx $$
This expression appears in the integrand, so we can substitute it with \( dt \):
$$ \int \frac{1}{x \sqrt{2x-1}} dx $$
$$ \int \frac{1}{x} \cdot \left( \frac{1}{\sqrt{2x-1}} dx \right) $$
$$ \int \frac{1}{x} \cdot dt $$
$$ \int \frac{1}{x} \ dt $$
Now we need to rewrite \( x \) in terms of \( t \). Starting from
$$ t = \sqrt{2x-1} $$
we square both sides:
$$ t^2 = 2x - 1 $$
Solving for \( x \):
$$ x = \frac{t^2 + 1}{2} $$
Substituting this back into the integral gives:
$$ \int \frac{1}{\frac{t^2 + 1}{2}} \ dt $$
$$ \int \frac{2}{t^2 + 1} \ dt $$
$$ 2 \int \frac{1}{t^2 + 1} \ dt $$
By applying the constant multiple rule, we factor out the 2:
$$ 2 \int \frac{1}{t^2 + 1} \ dt $$
This is a standard result: the antiderivative is the arctangent function.
$$ 2 \cdot \arctan(t) + C $$
Note. The derivative of arctangent is given by $$ D[\arctan t] = \frac{1}{1+t^2} $$
Since \( C \) is an arbitrary constant, writing \( 2C \) or \( C \) is equivalent, so we keep the simpler form:
$$ 2 \arctan(t) + C $$
Finally, we substitute back \( t = \sqrt{2x - 1} \):
$$ 2 \arctan(\sqrt{2x-1}) + C $$
This is the final result.
Method 2
Once again, the complication lies in the square root term:
$$ \int \frac{1}{x \sqrt{2x-1}} dx $$
We set:
$$ t = \sqrt{2x - 1} $$
Then solve for \( x \):
$$ t^2 = 2x - 1 \Rightarrow x = \frac{t^2 + 1}{2} $$
Now compute the differential of both sides:
$$ dx = \frac{d}{dt} \left( \frac{t^2 + 1}{2} \right) dt = \frac{2t}{2} dt = t \, dt $$
We now substitute \( dx = t \, dt \) into the original integral:
$$ \int \frac{1}{x \sqrt{2x-1}} dx = \int \frac{1}{x \cdot \sqrt{2x-1}} \cdot t \ dt $$
Recall that \( x = \frac{t^2 + 1}{2} \). Substituting, we get:
$$ \int \frac{1}{\frac{t^2 + 1}{2} \cdot \sqrt{2 \cdot \frac{t^2 + 1}{2} - 1}} \cdot t \ dt $$
Simplifying inside the square root:
$$ \sqrt{(t^2 + 1) - 1} = \sqrt{t^2} = t $$
So the integrand becomes:
$$ \int \frac{1}{\frac{t^2 + 1}{2} \cdot t} \cdot t \ dt $$
The \( t \)'s cancel out:
$$ \int \frac{1}{\frac{t^2 + 1}{2}} \ dt $$
Which simplifies to:
$$ \int \frac{2}{t^2 + 1} \ dt = 2 \int \frac{1}{t^2 + 1} \ dt $$
Again, the result is immediate:
$$ 2 \arctan(t) + C $$
And substituting back \( t = \sqrt{2x - 1} \):
$$ 2 \arctan(\sqrt{2x - 1}) + C $$
This is the final solution.
And that completes the exercise.
