Question

please answer this question​

Answer 1

Answer:

$\displaystyle \int {x^{-11}(1 + x^4)^\Big{- \frac{1}{2}}} \, dx = \boxed{ - \frac{\sqrt{x^4 + 1} (8x^8 - 4x^4 + 3)}{30x^{10}} + C }$

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Rule [Basic Power Rule]:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• Integrals

Integration Rule [Reverse Power Rule]:
$\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Property [Multiplied Constant]:
$\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]:
$\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Integration Methods: U-Substitution, U-Solve, Trigonometric Substitution

Step-by-step explanation:

*Note:

The problem is too big to fit all work. I will assume that you know how to do basic calculus.

Step 1: Define

Identify given.

$\displaystyle \int {x^{-11}(1 + x^4)^\Big{- \frac{1}{2}}} \, dx$

Step 2: Integrate Pt. 1

1. [Integrand] Rewrite:
   $\displaystyle \int {x^{-11}(1 + x^4)^\Big{- \frac{1}{2}}} \, dx = \int {\frac{1}{x^{11}\sqrt{x^4 + 1}}} \, dx$

Step 3: Integrate Pt. 2

Identify variables for u-substitution/u-solve.

1. Set u:
   $\displaystyle u = x^2$
2. [u] Differentiate [Derivative Rule - Basic Power Rule]:
   $\displaystyle du = 2x \ dx$
3. [du] Rewrite [U-Solve]:
   $\displaystyle dx = \frac{1}{2x} \ du$

Step 4: Integrate Pt. 3

Start solving the integral using u-solve:

$\displaystyle \begin{aligned}\int {x^{-11}(1 + x^4)^\Big{- \frac{1}{2}}} \, dx & = \int {\frac{1}{x^{11}\sqrt{x^4 + 1}}} \, dx \\& = \int {\frac{1}{2u^6 \sqrt{u^2 + 1}}} \, du \\& = \frac{1}{2} \int {\frac{1}{u^6 \sqrt{u^2 + 1}}} \, du \\\end{aligned}$

Step 5: Integrate Pt. 4

Identify variables for trigonometric substitution.

1. Set u:
   $\displaystyle\begin{aligned}u & = \tan v \\v & = \arctan u \\\end{aligned}$
2. [u] Differentiate [Triognometric Differentiation]:
   $\displaystyle du = \sec^2 v \ dv$

Step 6: Integrate Pt. 5

Let's focus on just the integral itself. Apply the Trigonometric Substitution Integration Method and other basic integration techniques listed under "Calculus":

$\displaystyle \begin{aligned}\int {\frac{1}{u^6 \sqrt{u^2 + 1} }} \, du & = \int {\frac{\sec^2 v}{\tan^6 v \sqrt{\tan^2 v + 1} }} \, dv \\& = \int {\frac{\sec v}{\tan^6 v}} \, dv \\& = \int {\cot v \csc v (\csc^2 v - 1)^2} \, dv \\\end{aligned}$

Step 7: Integrate Pt. 6

Identify variables for u-substitution/u-solve again.

Use another variable besides u to avoid confusion with earlier substitutions:

1. Set w:
   $\displaystyle w = \csc v$
2. [w] Differentiate [Triognometric Differentiation]:
   $\displaystyle dw = - \cot v \csc v \ dv$
3. [dw] Rewrite [U-Solve]:
   $\displaystyle dv = - \frac{1}{\cot v \csc v} \ dw$

Step 8: Integrate Pt. 7

Reduce the integral using the U-Solve Integration Method and other basic integration techniques listed under "Calculus":

$\displaystyle \begin{aligned}\int {\frac{1}{u^6 \sqrt{u^2 + 1} }} \, du & = \int {\frac{\sec^2 v}{\tan^6 v \sqrt{\tan^2 v + 1} }} \, dv \\& = \int {\frac{\sec v}{\tan^6 v}} \, dv \\& = \int {\cot v \csc v (\csc^2 v - 1)^2} \, dv \\& = \int {-(w^2 - 1)^2} \, dw \\& = - \int {(w^2 - 1)^2} \, dw \\& = - \int {\bigg( w^4 - 2w^2 + 1 \bigg)} \, dw \\\end{aligned}$

Step 9: Integrate Pt. 8

Solve the integral using basic integration techniques listed under "Calculus":

$\displaystyle\begin{aligned}- \int {\bigg( w^4 - 2w^2 + 1 \bigg)} \, dw & = - \Bigg[ \int {w^4} \, dx - \int {2w^2} \, dx + \int {} \, dx \Bigg] \\& = - \Bigg[ \int {w^4} \, dx - 2 \int {w^2} \, dx + \int {} \, dx \Bigg] \\& = - \Bigg[ \frac{w^5}{5} - \frac{2w^3}{3} + w + C \Bigg] \\& = - \frac{w^5}{5} + \frac{2w^3}{3} - w + C \\& = - \frac{\csc^5 v}{5} + \frac{2\csc^3 v}{3} - \csc v + C \\\end{aligned}$

$\displaystyle\begin{aligned}- \int {\bigg( w^4 - 2w^2 + 1 \bigg)} \, dw & = - \frac{(u^2 + 1)^\Big{\frac{5}{2}}}{5u^5} + \frac{2(u^2 + 1)^\Big{\frac{3}{2}}}{3u^3} - \frac{\sqrt{u^2 + 1}}{u} + C \\\end{aligned}$

Step 10: Integrate Pt. 9

Let's substitute our integral value into our integral from "Step 4":

$\displaystyle\begin{aligned}\frac{1}{2} \int {\frac{1}{u^6 \sqrt{u^2 + 1}}} \, du & = - \frac{(u^2 + 1)^\Big{\frac{5}{2}}}{10u^5} + \frac{(u^2 + 1)^\Big{\frac{3}{2}}}{3u^3} - \frac{\sqrt{u^2 + 1}}{2u} + C \\& = - \frac{(x^4 + 1)^\Big{\frac{5}{2}}}{10x^{10}} + \frac{(x^4 + 1)^\Big{\frac{3}{2}}}{3x^6} - \frac{\sqrt{x^4 + 1}}{2x^2} + C \\& = \boxed{ - \frac{\sqrt{x^4 + 1}(8x^8 - 4x^4 + 3)}{30x^{10}} + C } \\\end{aligned}$

∴ we have found the indefinite integral.

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Topic: Calculus