"Find the indefinite integral and check the result..." was answered by our experts

druczekq4

Answered question

2021-11-10

Find the indefinite integral and check the result by differentiation.

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

Provere

Beginner2021-11-11Added 18 answers

Step 1
Consider the following integral:

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

Substitute

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

in the above integral:

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

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

= \frac{1}{4} \times \frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C

= \frac{1}{4} \times 2 \sqrt{u} + C

= \frac{\sqrt{u}}{2} + C

Step 2
Substitute

u = 1 + x^{4}

in the above equation:

\int \frac{x^{3}}{\sqrt{1 + x^{4}}} d x = \frac{\sqrt{1 + x^{4}}}{2} + C

Differentiation part:

\frac{d}{d x} (\frac{\sqrt{1 + x^{4}}}{2} + C) = \frac{1}{2} \times \frac{1}{2 \sqrt{1 + x^{4}}} \times \frac{d}{d x} (1 + x^{4})

= \frac{1}{4 \sqrt{1 + x^{4}}} \times 4 x^{3}

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

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?