Question: "Evaluate the integral. ∫x4+3x2+1x5+5x3+5xdx"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

untchick04tm

Answered question

2021-12-09

Evaluate the integral.

\int \frac{x^{4} + 3 x^{2} + 1}{x^{5} + 5 x^{3} + 5 x} d x

Answer & Explanation

Timothy Wolff

Beginner2021-12-10Added 26 answers

Step 1
Given: $\int \frac{x^{4} + 3 x^{2} + 1}{x^{5} + 5 x^{3} + 5 x} d x$
for evaluating given integral we substitute
$x^{5} + 5 x^{3} + 5 x = t$
now differentiating both side with respect to x
$5 x^{4} + 5 (3 x^{2}) + 5 = \frac{d t}{d x} (∵ \frac{d}{d x} (k x^{n}) = k (n x^{n - 1}))$
$5 (x^{4} + 3 x^{2} + 1) = \frac{d t}{d x}$
$(x^{4} + 3 x^{2} + 1) d x = \frac{d t}{5}$
now substitute $(x^{4} + 3 x^{2} + 1) d x$ with $\frac{d t}{5}$ and $(x^{5} + 5 x^{3} + 5 x)$ with t in given integral
Step 2
so,
$\int \frac{x^{4} + 3 x^{2} + 1}{x^{5} + 5 x^{3} + 5 x} d x = \int \frac{d t}{5}$
$= \frac{1}{5} \int \frac{d t}{t} (∵ \int \frac{d x}{x} = \ln x + c)$
$= \frac{\ln t}{5} + c$
now replacing t with $(x^{5} + 5 x^{3} + 5 x)$
so, given integral is $\frac{\ln (x^{5} + 5 x^{3} + 5 x)}{5} + c$
hence, given integral is equal to $\frac{\ln (x^{5} + 5 x^{3} + 5 x)}{5} + c$ .

Deufemiak7

Beginner2021-12-11Added 34 answers

$\int \frac{x^{4} + 3 x^{2} + 1}{x^{5} + 5 x^{3} + 5 x} d x$
Transform the expression
$\int \frac{1}{5 t} d t$
Use properties of integrals
$\frac{1}{5} \times \int \frac{1}{t} d t$
Evaluate the integral
$\frac{1}{5} \times \ln (| t |)$
Substitute back
$\frac{1}{5} \times \ln (| x^{5} + 5 x^{3} + 5 x |)$
Add $C \in R$
Solution
$\frac{1}{5} \times \ln (| x^{5} + 5 x^{3} + 5 x |) + C$

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