What is the integral of sqrt(7+\sqrt{x)} ?

The integral of sqrt(7+\sqrt{x)} is 2(2/5 (7+sqrt(x))^{5/2}-14/3 (7+sqrt(x))^{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 sqrt(7+\sqrt{x)}

\int 7 + x d x

2 (\frac{2}{5} (7 + x)^{\frac{5}{2}} - \frac{14}{3} (7 + x)^{\frac{3}{2}}) + C

Solution steps

\int 7 + x d x

Apply u-substitution

: \int 7 + u \cdot 2 u d u

= \int 7 + u \cdot 2 u d u

Take the constant out:

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

= 2 \cdot \int 7 + u u d u

Apply u-substitution

: \int (v - 7) v d v

= 2 \cdot \int (v - 7) v d v

Expand

(v - 7) v : v^{\frac{3}{2}} - 7 v

= 2 \cdot \int v^{\frac{3}{2}} - 7 v d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int v^{\frac{3}{2}} d v - \int 7 v d v)

\int v^{\frac{3}{2}} d v = \frac{2}{5} v^{\frac{5}{2}}

\int 7 v d v = \frac{14}{3} v^{\frac{3}{2}}

= 2 (\frac{2}{5} v^{\frac{5}{2}} - \frac{14}{3} v^{\frac{3}{2}})

Substitute back

= 2 (\frac{2}{5} (7 + x)^{\frac{5}{2}} - \frac{14}{3} (7 + x)^{\frac{3}{2}})

Add a constant to the solution

= 2 (\frac{2}{5} (7 + x)^{\frac{5}{2}} - \frac{14}{3} (7 + x)^{\frac{3}{2}}) + C

\int \frac{sin ^{2} ( x ) + sin ( x )}{cos ^{2} ( x )} d x

\int (- 3 cos (x)) d x

\int - \frac{2}{x ^{2} + 2} d x

\int tanh (a x) d x

\int 5 x e^{3} d x

What is the integral of sqrt(7+\sqrt{x)} ?
The integral of sqrt(7+\sqrt{x)} is 2(2/5 (7+sqrt(x))^{5/2}-14/3 (7+sqrt(x))^{3/2})+C