Solved: "Find the indefinite integrals (most general antiderivatives). You..."

Mylo O'Moore

Answered question

2021-10-23

Find the indefinite integrals (most general antiderivatives). You may need to try a solution and then adjust your guess. Check your answers by differentiation.

\int x^{3} {(1 + x^{4})}^{\frac{- 1}{4}} d x

Answer & Explanation

wheezym

Skilled2021-10-24Added 103 answers

Solution:

I = \int x^{3} {(1 + x^{4})}^{\frac{- 1}{4}} d x

I = \int \frac{x^{3}}{\sqrt{4} {x^{4} + 1}} d x

Let

u = x^{4} + 1

Then

d u = {(x^{4} + 1)}^{'} d x = 4 x^{3} d x

and we have that

x^{3} d x = \frac{d u}{4}

.
So,

\int \frac{x^{3}}{\sqrt{4} {x^{4} + 1}} d x = \int \frac{1}{4 \sqrt{4} {u}} d u

Apply the constant multiple rule

\int c f (u) d u = c \int f (u) d u

with

c = \frac{1}{4}

and

f (u) = \frac{1}{\sqrt{4} {u}}

\int \frac{1}{4 \sqrt{4} {u}} d u = (\frac{\int \frac{1}{\sqrt{4} {u}} d u}{4})

Apply the power rule

\int u^{n} d u = \frac{u^{n + 1}}{n + 1} (n \neq - 1)

with

n = - \frac{1}{4}

\frac{\int \frac{1}{\sqrt{4} {u} d u}}{4} = \frac{\int u^{- \frac{1}{4}}}{4} = \frac{\frac{u^{- \frac{1}{4} + 1}}{- \frac{1}{4} + 1}}{4} = \frac{(\frac{4 u^{\frac{3}{4}}}{3})}{4}

Recall that

u = x^{4} + 1

\frac{u^{\frac{3}{4}}}{3} = \frac{{(x^{4} + 1)}^{\frac{3}{4}}}{3}

Therefore,

\int \frac{x^{3}}{\sqrt{4} {}

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