What is the integral of (-2sec^2(x))/((4+tan(x))^3) ?

The integral of (-2sec^2(x))/((4+tan(x))^3) is 1/((4+tan(x))^2)+C

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integral of (-2sec^2(x))/((4+tan(x))^3)

\int \frac{- 2 sec ^{2} ( x )}{( 4 + tan ( x ) ) ^{3}} d x

\frac{1}{( 4 + tan ( x ) ) ^{2}} + C

Solution steps

\int \frac{- 2 sec ^{2} ( x )}{( 4 + tan ( x ) ) ^{3}} d x

Take the constant out:

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

= - 2 \cdot \int \frac{sec ^{2} ( x )}{( 4 + tan ( x ) ) ^{3}} d x

Apply u-substitution

= - 2 \cdot \int \frac{1}{( 4 + u ) ^{3}} d u

Apply u-substitution

= - 2 \cdot \int \frac{1}{v ^{3}} d v

Apply the Power Rule

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

Substitute back

= - 2 (- \frac{1}{2 ( 4 + tan ( x ) ) ^{2}})

Simplify

- 2 (- \frac{1}{2 ( 4 + tan ( x ) ) ^{2}}) : \frac{1}{( 4 + tan ( x ) ) ^{2}}

= \frac{1}{( 4 + tan ( x ) ) ^{2}}

Add a constant to the solution

= \frac{1}{( 4 + tan ( x ) ) ^{2}} + C

What is the integral of (-2sec^2(x))/((4+tan(x))^3) ?
The integral of (-2sec^2(x))/((4+tan(x))^3) is 1/((4+tan(x))^2)+C