What is the integral of (4e^x)/(e^{2x)+10e^x+25} ?

The integral of (4e^x)/(e^{2x)+10e^x+25} is -4/(e^x+5)+C

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integral of (4e^x)/(e^{2x)+10e^x+25}

\int \frac{4 e ^{x}}{e ^{2 x} + 10 e ^{x} + 25} d x

- \frac{4}{e ^{x} + 5} + C

Solution steps

\int \frac{4 e ^{x}}{e ^{2 x} + 10 e ^{x} + 25} d x

Take the constant out:

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

= 4 \cdot \int \frac{e ^{x}}{e ^{2 x} + 10 e ^{x} + 25} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 10 u + 25} d u

= 4 \cdot \int \frac{1}{u ^{2} + 10 u + 25} d u

Take the partial fraction of

\frac{1}{u ^{2} + 10 u + 25} : \frac{1}{( u + 5 ) ^{2}}

= 4 \cdot \int \frac{1}{( u + 5 ) ^{2}} d u

Apply u-substitution

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

= 4 \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 4 (- \frac{1}{v})

Substitute back

= 4 (- \frac{1}{e ^{x} + 5})

Simplify

4 (- \frac{1}{e ^{x} + 5}) : - \frac{4}{e ^{x} + 5}

= - \frac{4}{e ^{x} + 5}

Add a constant to the solution

= - \frac{4}{e ^{x} + 5} + C

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

y^{^{'}} = (y + 2 t)^{2}

- 2 y - y^{2} d x + (3 + 2 x - x^{2}) d y = 0

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

\frac{d}{d x} (- 2 a sin (2 x) + 2 b cos (2 x))

What is the integral of (4e^x)/(e^{2x)+10e^x+25} ?
The integral of (4e^x)/(e^{2x)+10e^x+25} is -4/(e^x+5)+C