What is the integral from 1 to x of 5/(t^3+7) ?

The integral from 1 to x of 5/(t^3+7) is (5x-5)/(7+t^3)

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integral from 1 to x of 5/(t^3+7)

\int_{1}^{x} \frac{5}{t ^{3} + 7} d x

\frac{5 x - 5}{7 + t ^{3}}

Solution steps

\int_{1}^{x} \frac{5}{t ^{3} + 7} d x

Integral of a constant:

\int a d x = a x

= [\frac{5}{7 + t ^{3}} x]_{1}^{x}

Compute the boundaries:

\frac{5}{7 + t ^{3}} x - \frac{5}{7 + t ^{3}}

= \frac{5}{7 + t ^{3}} x - \frac{5}{7 + t ^{3}}

Simplify

= \frac{5 x - 5}{7 + t ^{3}}

\int_{0}^{3} ∣ 1 - x ∣ d x

\int_{- π}^{2 x} \frac{1}{1 + x ^{4}} d x

\int_{0}^{5} 1 + 2 x^{3} d x

\int_{0}^{3} 2 x (x - 7) d x

\int_{x}^{π} sin (y^{2}) d y

What is the integral from 1 to x of 5/(t^3+7) ?
The integral from 1 to x of 5/(t^3+7) is (5x-5)/(7+t^3)