What is the integral from x to x^2 of y/x ?

The integral from x to x^2 of y/x is (x(x^2-1))/2

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integral from x to x^2 of y/x

\int_{x}^{x^{2}} \frac{y}{x} d y

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

Solution steps

\int_{x}^{x^{2}} \frac{y}{x} d y

Take the constant out:

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

= \frac{1}{x} \cdot \int_{x}^{x^{2}} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{x}^{x^{2}}

= \frac{1}{x} [\frac{y ^{2}}{2}]_{x}^{x^{2}}

Simplify

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

= \frac{[ \frac{y ^{2}}{2} ] _{x}^{x^{2}}}{x}

Compute the boundaries:

\frac{x ^{4}}{2} - \frac{x ^{2}}{2}

= \frac{\frac{x ^{4}}{2} - \frac{x ^{2}}{2}}{x}

Simplify

= \frac{x ( x ^{2} - 1 )}{2}

\int_{1}^{e} (x ln (x))^{2} d x

\int_{0}^{2} \frac{2}{1 - x ^{2}} d x

\int_{0}^{4} 6 2 x + 1 d x

\int_{1}^{6} ln (1 + x) d x

\int_{- 1}^{1} 12 e^{- 3 x} d x

What is the integral from x to x^2 of y/x ?
The integral from x to x^2 of y/x is (x(x^2-1))/2