What is the integral of (5x+1)^4 ?

The integral of (5x+1)^4 is 1/25 (5x+1)^5+C

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integral of (5x+1)^4

\int (5 x + 1)^{4} d x

\frac{1}{25} (5 x + 1)^{5} + C

Solution steps

\int (5 x + 1)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{5} d u

= \int \frac{u ^{4}}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{1}{5} \cdot \frac{u ^{5}}{5}

Substitute back

u = 5 x + 1

= \frac{1}{5} \cdot \frac{( 5 x + 1 ) ^{5}}{5}

Simplify

\frac{1}{5} \cdot \frac{( 5 x + 1 ) ^{5}}{5} : \frac{1}{25} (5 x + 1)^{5}

= \frac{1}{25} (5 x + 1)^{5}

Add a constant to the solution

= \frac{1}{25} (5 x + 1)^{5} + C

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

f (x) = 9 - ∣ x - 4 ∣ at x

x \to 5 - lim (\frac{2}{x ^{2} - 25})

\int_{- 2}^{1} (12 x^{5} - 36) d x

x^{2} y^{^{''}} + x y^{^{'}} - y = 0, y (1) = 1, y^{^{'}} (1) = 4