Sure, let's go through the indefinite integral step by step using the substitution [tex]\( u = 2x^2 + 3x \)[/tex].
First, we substitute [tex]\( u = 2x^2 + 3x \)[/tex], which means we need to compute the differential [tex]\( du \)[/tex] in terms of [tex]\( dx \)[/tex].
1. Find [tex]\( du \)[/tex] in terms of [tex]\( dx \)[/tex]:
Differentiate [tex]\( u = 2x^2 + 3x \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
\frac{du}{dx} = 4x + 3
\][/tex]
Hence,
[tex]\[
du = (4x + 3) \, dx
\][/tex]
Solving for [tex]\( dx \)[/tex], we get:
[tex]\[
dx = \frac{du}{4x + 3}
\][/tex]
2. Substitute [tex]\( u \)[/tex] and [tex]\( dx \)[/tex] back into the original integral:
The given integral is:
[tex]\[
\int (4x + 3) \sqrt{2x^2 + 3x} \, dx
\][/tex]
Substitute [tex]\( u = 2x^2 + 3x \)[/tex] and [tex]\( dx = \frac{du}{4x + 3} \)[/tex]:
[tex]\[
\int (4x + 3) \sqrt{u} \, \frac{du}{4x + 3}
\][/tex]
Notice the [tex]\( 4x + 3 \)[/tex] terms will cancel each other out:
[tex]\[
\int \sqrt{u} \, du
\][/tex]
3. Write the integrand in terms of [tex]\( u \)[/tex]:
Finally, the rewritten integral in terms of [tex]\( u \)[/tex] is:
[tex]\[
\int (4x + 3) \sqrt{2x^2 + 3x} \, dx = \int \sqrt{u} \, du
\][/tex]
So, when using the substitution [tex]\( u = 2x^2 + 3x \)[/tex], the integral [tex]\(\int(4x+3) \sqrt{2x^2+3x} \, dx\)[/tex] transforms into:
[tex]\[
\int \sqrt{u} \, du
\][/tex]
Now, you can evaluate the integral [tex]\(\int \sqrt{u} \, du \)[/tex] to find the final result.
