What is the integral from-1 to 1 of t^3(1+t^4)^3 ?

The integral from-1 to 1 of t^3(1+t^4)^3 is 0

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integral from-1 to 1 of t^3(1+t^4)^3

\int_{- 1}^{1} t^{3} (1 + t^{4})^{3} d t

0

Solution steps

\int_{- 1}^{1} t^{3} (1 + t^{4})^{3} d t

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

[- a, a]

then

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

Parity of

t^{3} (1 + t^{4})^{3} :

Odd

= 0

\frac{d}{d x} (\frac{e ^{x}}{ln ( x )})

\int 1 - \frac{x}{2} d x

\int \frac{ln ( x )}{2 x ^{2}} d x

5 (- csc^{2} (x) + sec^{2} (x))

\frac{4}{u}

What is the integral from-1 to 1 of t^3(1+t^4)^3 ?
The integral from-1 to 1 of t^3(1+t^4)^3 is 0