What is the integral from 0 to 9 of (10)/((1+x)^2) ?

The integral from 0 to 9 of (10)/((1+x)^2) is 9

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integral from 0 to 9 of (10)/((1+x)^2)

\int_{0}^{9} \frac{10}{( 1 + x ) ^{2}} d x

9

Solution steps

\int_{0}^{9} \frac{10}{( 1 + x ) ^{2}} d x

Take the constant out:

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

= 10 \cdot \int_{0}^{9} \frac{1}{( 1 + x ) ^{2}} d x

Apply u-substitution

: \int_{1}^{10} \frac{1}{u ^{2}} d u

= 10 \cdot \int_{1}^{10} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{1}^{10}

= 10 [- \frac{1}{u}]_{1}^{10}

Compute the boundaries:

\frac{9}{10}

= 10 \cdot \frac{9}{10}

Simplify

= 9

x \to - 3 lim (ln (x + 3))

cos (x^{2} - 4 x + 3)

\int \frac{t}{e ^{2 t}} d t

\int_{- 2}^{9} f (x) d x

\frac{\partial}{\partial y} (2 x y^{3} + 3)

What is the integral from 0 to 9 of (10)/((1+x)^2) ?
The integral from 0 to 9 of (10)/((1+x)^2) is 9