What is the integral of (x/4)^2 ?

The integral of (x/4)^2 is (x^3)/(16)-(x^3)/(24)+C

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integral of (x/4)^2

\int (\frac{x}{4})^{2} d x

\frac{x ^{3}}{16} - \frac{x ^{3}}{24} + C

Solution steps

\int (\frac{x}{4})^{2} d x

Apply Integration By Parts

= \frac{x ^{3}}{16} - \int \frac{x ^{2}}{8} d x

\int \frac{x ^{2}}{8} d x = \frac{x ^{3}}{24}

= \frac{x ^{3}}{16} - \frac{x ^{3}}{24}

Add a constant to the solution

= \frac{x ^{3}}{16} - \frac{x ^{3}}{24} + C

\int \frac{t}{t} d t

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

in t e g r a l \frac{1}{2 ^{x}}

\int (cos^{5} (x)) d x

\int \frac{7 x ^{2} + x + 12}{3 x ^{3} + 12 x} d x

What is the integral of (x/4)^2 ?
The integral of (x/4)^2 is (x^3)/(16)-(x^3)/(24)+C