What is the integral from-1 to 7 of (x^2)/(x^3+1) ?

The integral from-1 to 7 of (x^2)/(x^3+1) is diverges

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integral from-1 to 7 of (x^2)/(x^3+1)

\int_{- 1}^{7} \frac{x ^{2}}{x ^{3} + 1} d x

diverges

Solution steps

\int_{- 1}^{7} \frac{x ^{2}}{x ^{3} + 1} d x

Apply u-substitution

: \int_{0}^{344} \frac{1}{3 u} d u

= \int_{0}^{344} \frac{1}{3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{0}^{344} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{0}^{344}

Compute the boundaries:

diverges

= diverges

\int_{- 2}^{0} 4 - 2^{- x} d x

\int_{- 3}^{3} x^{2} - 1 d x

\int_{0}^{1} 81 x^{2} + 729 d x

\int_{- 3}^{2} \frac{1}{( x + 1 ) ^{3}} d x

\int_{6}^{2} x^{3} d x

What is the integral from-1 to 7 of (x^2)/(x^3+1) ?
The integral from-1 to 7 of (x^2)/(x^3+1) is diverges