What is the integral of e^{5x^2+10x+10}(x+1) ?

The integral of e^{5x^2+10x+10}(x+1) is 1/10 e^{10+5x^2+10x}+C

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integral of e^{5x^2+10x+10}(x+1)

\int e^{5 x^{2} + 10 x + 10} (x + 1) d x

\frac{1}{10} e^{10 + 5 x^{2} + 10 x} + C

Solution steps

\int e^{5 x^{2} + 10 x + 10} (x + 1) d x

e^{5 x^{2} + 10 x + 10} (x + 1) = e^{10} e^{5 x^{2} + 10 x} (x + 1)

= \int e^{10} e^{5 x^{2} + 10 x} (x + 1) d x

Take the constant out:

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

= e^{10} \cdot \int e^{5 x^{2} + 10 x} (x + 1) d x

Apply u-substitution

: \int \frac{e ^{u}}{10} d u

= e^{10} \cdot \int \frac{e ^{u}}{10} d u

Take the constant out:

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

= e^{10} \frac{1}{10} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= e^{10} \frac{1}{10} e^{u}

Substitute back

u = 5 x^{2} + 10 x

= e^{10} \frac{1}{10} e^{5 x^{2} + 10 x}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e^{10} e^{5 x^{2} + 10 x} = e^{10 + 5 x^{2} + 10 x}

= \frac{1}{10} e^{10 + 5 x^{2} + 10 x}

Add a constant to the solution

= \frac{1}{10} e^{10 + 5 x^{2} + 10 x} + C

\frac{d}{d x} (sin (\frac{x}{8}))

x \to 2 π - lim (e^{\frac{1}{x}} - x)

x \to - \frac{π}{2} lim (csc (\frac{x}{2}))

ln (1 + 2 x)

y = x 5^{2 x}

What is the integral of e^{5x^2+10x+10}(x+1) ?
The integral of e^{5x^2+10x+10}(x+1) is 1/10 e^{10+5x^2+10x}+C