What is the integral of 1/(sqrt(h-16)) ?

The integral of 1/(sqrt(h-16)) is 2sqrt(h-16)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(h-16))

\int \frac{1}{h - 16} d h

2 h - 16 + C

Solution steps

\int \frac{1}{h - 16} d h

Apply u-substitution

: \int 2 d u

= \int 2 d u

Integral of a constant:

\int a d x = a x

= 2 u

Substitute back

u = h - 16

= 2 h - 16

Add a constant to the solution

= 2 h - 16 + C

\int \frac{5 x ^{2}}{3 x ^{3} + 7} d x

\int (3 x^{2} + 4) d x

\int \frac{t ^{3}}{t ^{2} - 4 t + 4} d t

\int 12 x ln (8 x) d x

\int 9 - (x - 1)^{2} d x

What is the integral of 1/(sqrt(h-16)) ?
The integral of 1/(sqrt(h-16)) is 2sqrt(h-16)+C