What is the integral of 9xsqrt(1+2x^2) ?

The integral of 9xsqrt(1+2x^2) is 3/2 (2x^2+1)^{3/2}+C

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integral of 9xsqrt(1+2x^2)

\int 9 x 1 + 2 x^{2} d x

\frac{3}{2} (2 x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int 9 x 1 + 2 x^{2} d x

Take the constant out:

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

= 9 \cdot \int x 1 + 2 x^{2} d x

Apply u-substitution

: \int \frac{u}{4} d u

= 9 \cdot \int \frac{u}{4} d u

Take the constant out:

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

= 9 \cdot \frac{1}{4} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 9 \cdot \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 2 x^{2}

= 9 \cdot \frac{1}{4} \cdot \frac{2}{3} (1 + 2 x^{2})^{\frac{3}{2}}

Simplify

9 \cdot \frac{1}{4} \cdot \frac{2}{3} (1 + 2 x^{2})^{\frac{3}{2}} : \frac{3}{2} (2 x^{2} + 1)^{\frac{3}{2}}

= \frac{3}{2} (2 x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{3}{2} (2 x^{2} + 1)^{\frac{3}{2}} + C

n \to \infty lim (0)

7 x - 2

\frac{d}{d x} ((x - 2)^{2} (x + 4))

(4 x (- 9)^{^{'}})^{^{'}}

4 x \cdot ln (x)

What is the integral of 9xsqrt(1+2x^2) ?
The integral of 9xsqrt(1+2x^2) is 3/2 (2x^2+1)^{3/2}+C