What is the integral from-2 to 2 of xln(1+x^4) ?

The integral from-2 to 2 of xln(1+x^4) is 0

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integral from-2 to 2 of xln(1+x^4)

\int_{- 2}^{2} x ln (1 + x^{4}) d x

0

Solution steps

\int_{- 2}^{2} x ln (1 + x^{4}) d x

If f(x) is an odd function and continuous at:

[- a, a]

then

\int_{- a}^{a} f (x) d x = 0

Parity of

x ln (1 + x^{4}) :

Odd

= 0

\int_{0}^{2 π} sin (t) cos (t) d t

\int_{0}^{x} \frac{x}{2} d x

\int_{\frac{- 1}{8}}^{\frac{1}{8}} sin (2 π x) d x

\int_{0}^{π} x \cdot cos (n x) d x

\int_{0}^{2} x cos (\frac{nπ x}{2}) d x

What is the integral from-2 to 2 of xln(1+x^4) ?
The integral from-2 to 2 of xln(1+x^4) is 0