What is the integral of xsqrt(32+4x^2) ?

The integral of xsqrt(32+4x^2) is 2/3 (8+x^2)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of xsqrt(32+4x^2)

\int x 32 + 4 x^{2} d x

\frac{2}{3} (8 + x^{2})^{\frac{3}{2}} + C

Solution steps

\int x 32 + 4 x^{2} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 32 + 4 x^{2}

= \frac{1}{8} \cdot \frac{2}{3} (32 + 4 x^{2})^{\frac{3}{2}}

Simplify

\frac{1}{8} \cdot \frac{2}{3} (32 + 4 x^{2})^{\frac{3}{2}} : \frac{2}{3} (8 + x^{2})^{\frac{3}{2}}

= \frac{2}{3} (8 + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (8 + x^{2})^{\frac{3}{2}} + C

\int 0.2 x y d x

\int e^{2 x} \cdot (2 x^{2} + 8 x + 3) d x

\int (12 x^{3} + 6 x - 2) d x

\int \frac{3}{4 x ^{2}} d x

\int \frac{e ^{8 x}}{e ^{8 x} + 3} d x

What is the integral of xsqrt(32+4x^2) ?
The integral of xsqrt(32+4x^2) is 2/3 (8+x^2)^{3/2}+C