What is the integral of 8xsqrt(4x^2+5) ?

The integral of 8xsqrt(4x^2+5) is 2/3 (4x^2+5)^{3/2}+C

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integral of 8xsqrt(4x^2+5)

\int 8 x 4 x^{2} + 5 d x

\frac{2}{3} (4 x^{2} + 5)^{\frac{3}{2}} + C

Solution steps

\int 8 x 4 x^{2} + 5 d x

Take the constant out:

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

= 8 \cdot \int x 4 x^{2} + 5 d x

Apply u-substitution

: \int \frac{u}{8} d u

= 8 \cdot \int \frac{u}{8} d u

Take the constant out:

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

= 8 \cdot \frac{1}{8} \cdot \int u d u

Apply the Power Rule

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

= 8 \cdot \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 x^{2} + 5

= 8 \cdot \frac{1}{8} \cdot \frac{2}{3} (4 x^{2} + 5)^{\frac{3}{2}}

Simplify

8 \cdot \frac{1}{8} \cdot \frac{2}{3} (4 x^{2} + 5)^{\frac{3}{2}} : \frac{2}{3} (4 x^{2} + 5)^{\frac{3}{2}}

= \frac{2}{3} (4 x^{2} + 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (4 x^{2} + 5)^{\frac{3}{2}} + C

x \to 1 lim (x + 8)

x y^{^{'}} + y = \frac{1}{y ^{2}}

\int x^{12} ln (x) d x

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + 2 y = 0

\int (2 + tan (x))^{2} d x

What is the integral of 8xsqrt(4x^2+5) ?
The integral of 8xsqrt(4x^2+5) is 2/3 (4x^2+5)^{3/2}+C