What is the integral of 5/((x+1)^2) ?

The integral of 5/((x+1)^2) is -5/(x+1)+C

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integral of 5/((x+1)^2)

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

- \frac{5}{x + 1} + C

Solution steps

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

Take the constant out:

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

= 5 \cdot \int \frac{1}{( x + 1 ) ^{2}} d x

Apply u-substitution

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

= 5 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 5 (- \frac{1}{u})

Substitute back

u = x + 1

= 5 (- \frac{1}{x + 1})

Simplify

5 (- \frac{1}{x + 1}) : - \frac{5}{x + 1}

= - \frac{5}{x + 1}

Add a constant to the solution

= - \frac{5}{x + 1} + C

f (t) = t 4 - t^{2}

(y^{2} + 2 x y) d x - x^{2} d y = 0

\frac{d y}{d x} = \frac{3 x ^{2} + 4 x + 2}{2 ( y - 1 )}, y (0) = - 1

y^{^{'}} = sin^{2} (x - y + 1)

\frac{d y}{d x} + 2 y = x