Resolved: "Evaluate the indefinite integral. ∫z2(z3+1)12dz"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

Algotssleeddynf

Answered question

2021-12-09

Evaluate the indefinite integral.

\int z^{2} {(z^{3} + 1)}^{12} d z

Answer & Explanation

Charles Benedict

Beginner2021-12-10Added 32 answers

Step 1
We have evaluate $\int z^{2} {(z^{3} + 1)}^{12} d z$
Let substitute $t = z^{3} + 1$
Differentiating both both sides with respect to z,
$\frac{d t}{d z} = 3 z^{2} + 0$
$\frac{d t}{3} = z^{2} d z$
Substitute the above value in given integral,
$\int z^{2} {(z^{3} + 1)}^{12} d z = \int {(z^{3} + 1)}^{12} z^{2} d z$
$= \int t^{12} \cdot \frac{d t}{3}$
$= \frac{1}{3} \int t^{12} d t$
$= \frac{1}{3} \cdot \frac{t^{12 + 1}}{12 + 1} + C$
$= \frac{1}{39} t^{13} + C$
Step 2
Put back $t = (z^{3} + 1)$
$\int z^{2} {(z^{3} + 1)}^{12} d z = \frac{1}{39} {(z^{3} + 1)}^{13} + C$
Hence, required integral is $\int z^{2} {(z^{3} + 1)}^{12} d z = \frac{1}{39} {(z^{3} + 1)}^{13} + C$ .

Jim Hunt

Beginner2021-12-11Added 45 answers

Given:
$\int z^{2} {(z^{3} + 1)}^{12} d z$
$\int \frac{t^{12}}{3} d t \frac{1}{3} * \int t^{12} d t$
$\frac{1}{3} \cdot \frac{t^{13}}{13}$
$\frac{1}{3} \cdot \frac{{(z^{3} + 1)}^{13}}{13}$
$\frac{{(z^{3} + 1)}^{13}}{39}$
Add C
Answer:
$\frac{{(z^{3} + 1)}^{13}}{39} + C$

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