Answered: "Use this only Power Formula Logarithm Formula Exponential..."

Maaghu

Answered question

2021-11-28

Use this only Power Formula Logarithm Formula Exponential Functions Trigonometric Functions. Complete solution pls

\int_{0}^{1} \frac{2 x^{3} + x^{2} + 3}{{(x^{2} + 1)}^{2}} d x

Answer & Explanation

Lounctirough

Beginner2021-11-29Added 14 answers

Given $\int_{0}^{1} \frac{2 x^{3} + x^{2} + 3}{{(x^{2} + 1)}^{2}} d x$
Perf orm partial decomposition:
$= \int \frac{2 x + 1}{x^{2} + 1} + \frac{2 - 2 x}{{(x^{2} + 1)}^{2}} d x$
Apply lienarity:
$= \int \frac{2 x + 1}{x^{2} + 1} d x - 2 \int \frac{x - 1}{{(x^{2} + 1)}^{2}} d x$
Now solve
$\int \frac{2 x + 1}{x^{2} + 1} d x$
Expand:
$= \int \frac{2 x}{x^{2} + 1} d x + \int \frac{1}{x^{2} + 1} d x$
Substitute $u = x^{2} + 1 \to \frac{d u}{d x} = 2 x \to d x = \frac{1}{2 x} d u$
$= \frac{1}{2} \int \frac{1}{u} d u$
We knov that $\int \frac{1}{x} d x = \ln (x)$
$= \frac{\ln (u)}{2}$
Undo the substitution:
$= \frac{\ln (x^{2} + 1)}{2}$
We know that $\int \frac{1}{x^{2} + 1} d x = \arctan (x)$
Plug in solved integrals:
$= \ln (x^{2} + 1) + ∣ \arctan (x)$
Now solvind: $\int \frac{x - 1}{{(x^{2} + 1)}^{2}}$
Expand:
$= \int \frac{x}{{(x^{2} + 1)}^{2} d x - \int \frac{1}{{(x^{2} + 1)}^{2} d x}}$
Substitute $u = x^{2} + 1 \to \frac{d u}{d x} = 2 x \to d x = \frac{1}{2 x} d u$

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