What is the integral of ((x^2+7)^{2021})x ?

The integral of ((x^2+7)^{2021})x is 1/4044 (x^2+7)^{2022}+C

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integral of ((x^2+7)^{2021})x

\int ((x^{2} + 7)^{2021}) x d x

\frac{1}{4044} (x^{2} + 7)^{2022} + C

Solution steps

\int (x^{2} + 7)^{2021} x d x

Apply u-substitution

: \int \frac{u ^{2021}}{2} d u

= \int \frac{u ^{2021}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{2021} d u

Apply the Power Rule

: \frac{u ^{2022}}{2022}

= \frac{1}{2} \cdot \frac{u ^{2022}}{2022}

Substitute back

u = x^{2} + 7

= \frac{1}{2} \cdot \frac{( x ^{2} + 7 ) ^{2022}}{2022}

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} + 7 ) ^{2022}}{2022} : \frac{1}{4044} (x^{2} + 7)^{2022}

= \frac{1}{4044} (x^{2} + 7)^{2022}

Add a constant to the solution

= \frac{1}{4044} (x^{2} + 7)^{2022} + C

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

\frac{d}{d x} ((x + (x + sin^{2} (x))^{3})^{4})

x \to - 4 lim (\frac{x ^{2}}{4 x + 16})

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

What is the integral of ((x^2+7)^{2021})x ?
The integral of ((x^2+7)^{2021})x is 1/4044 (x^2+7)^{2022}+C