What is the integral from 1 to t of 9/(x^3) ?

The integral from 1 to t of 9/(x^3) is 9(-1/(2t^2)+1/2)

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integral from 1 to t of 9/(x^3)

\int_{1}^{t} \frac{9}{x ^{3}} d x

9 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

Solution steps

\int_{1}^{t} \frac{9}{x ^{3}} d x

Take the constant out:

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

= 9 \cdot \int_{1}^{t} \frac{1}{x ^{3}} d x

Apply the Power Rule

: [- \frac{1}{2 x ^{2}}]_{1}^{t}

= 9 [- \frac{1}{2 x ^{2}}]_{1}^{t}

Compute the boundaries:

- \frac{1}{2 t ^{2}} + \frac{1}{2}

= 9 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

\frac{d}{d x} (\frac{5 x}{( x ^{2} + 1 ) ^{\frac{3}{5}}})

\frac{\partial}{\partial x} (sin (4 x + 4 y))

\frac{d}{d x} ((1 - x^{2})^{5})

f (x) = csc (x^{2})

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

What is the integral from 1 to t of 9/(x^3) ?
The integral from 1 to t of 9/(x^3) is 9(-1/(2t^2)+1/2)