What is the integral of (-30)/((2x+1)^2) ?

The integral of (-30)/((2x+1)^2) is (15)/(2x+1)+C

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integral of (-30)/((2x+1)^2)

\int \frac{- 30}{( 2 x + 1 ) ^{2}} d x

\frac{15}{2 x + 1} + C

Solution steps

\int \frac{- 30}{( 2 x + 1 ) ^{2}} d x

Take the constant out:

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

= - 30 \cdot \int \frac{1}{( 2 x + 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2}} d u

= - 30 \cdot \int \frac{1}{2 u ^{2}} d u

Take the constant out:

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

= - 30 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - 30 \cdot \frac{1}{2} (- \frac{1}{u})

Substitute back

u = 2 x + 1

= - 30 \cdot \frac{1}{2} (- \frac{1}{2 x + 1})

Simplify

- 30 \cdot \frac{1}{2} (- \frac{1}{2 x + 1}) : \frac{15}{2 x + 1}

= \frac{15}{2 x + 1}

Add a constant to the solution

= \frac{15}{2 x + 1} + C

\int \frac{- sin ( x )}{cos ^{2} ( x )} d x

x \to 1 - lim (x^{2} + 1)

\int \frac{1}{7 - 5 x ^{2}} d x

\frac{1}{1 - x}, i

y = x^{3}, y = x^{2} - 4 x, [- 1, 1]

What is the integral of (-30)/((2x+1)^2) ?
The integral of (-30)/((2x+1)^2) is (15)/(2x+1)+C