What is the integral of 6xsqrt(1+3^2) ?

The integral of 6xsqrt(1+3^2) is 3sqrt(10)x^2+C

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integral of 6xsqrt(1+3^2)

\int 6 x 1 + 3^{2} d x

310 x^{2} + C

Solution steps

\int 6 x 1 + 3^{2} d x

Take the constant out:

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

= 6 1 + 3^{2} \cdot \int x d x

Simplify

= 610 \cdot \int x d x

Apply the Power Rule

: \frac{x ^{2}}{2}

= 610 \frac{x ^{2}}{2}

Simplify

610 \frac{x ^{2}}{2} : 310 x^{2}

= 310 x^{2}

Add a constant to the solution

= 310 x^{2} + C

\frac{( ( x + 1 ) ( x + 2 ) )}{x}

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\int \frac{( 4 - ln ∣ x + 3 ∣ ) ^{3}}{x + 3} d x

\frac{d}{d x} (x^{2} \cdot e^{- x})

\int_{0}^{1} x 3^{x} d x

What is the integral of 6xsqrt(1+3^2) ?
The integral of 6xsqrt(1+3^2) is 3sqrt(10)x^2+C