What is the integral of sqrt(4x^2+7) ?

The integral of sqrt(4x^2+7) is 1/4 (2xsqrt(4x^2+7)+7(ln|2x+sqrt(7)sqrt((4x^2+7)/7)|-(ln(7))/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(4x^2+7)

\int 4 x^{2} + 7 d x

\frac{1}{4} (2 x 4 x^{2} + 7 + 7 (ln 2 x + 7 \frac{4 x ^{2} + 7}{7} - \frac{ln ( 7 )}{2})) + C

Solution steps

\int 4 x^{2} + 7 d x

Apply Trigonometric Substitution

: \int \frac{7}{2} sec^{3} (u) d u

= \int \frac{7}{2} sec^{3} (u) d u

Take the constant out:

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

= \frac{7}{2} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{7}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{7}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{2}{7} x)

= \frac{7}{2} \frac{sec ^{2} ( arctan ( \frac{2}{7} x ) ) sin ( arctan ( \frac{2}{7} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{7} x)) + sec (arctan (\frac{2}{7} x))

Simplify

\frac{7}{2} \frac{sec ^{2} ( arctan ( \frac{2}{7} x ) ) sin ( arctan ( \frac{2}{7} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{7} x)) + sec (arctan (\frac{2}{7} x)) : \frac{1}{4} (2 x 4 x^{2} + 7 + 7 (ln 2 x + 7 \frac{4 x ^{2} + 7}{7} - \frac{ln ( 7 )}{2}))

= \frac{1}{4} (2 x 4 x^{2} + 7 + 7 (ln 2 x + 7 \frac{4 x ^{2} + 7}{7} - \frac{ln ( 7 )}{2}))

Add a constant to the solution

= \frac{1}{4} (2 x 4 x^{2} + 7 + 7 (ln 2 x + 7 \frac{4 x ^{2} + 7}{7} - \frac{ln ( 7 )}{2})) + C

\frac{\partial}{\partial x} (4 x^{2} y^{3} - x z)

n = 1 \sum \infty \frac{1 + n}{3 ^{n}}

\int 8 sin^{2} (x) d x

y = t^{2} sin (t)

\int \frac{1}{3 x ^{2} + 2} d x

What is the integral of sqrt(4x^2+7) ?
The integral of sqrt(4x^2+7) is 1/4 (2xsqrt(4x^2+7)+7(ln|2x+sqrt(7)sqrt((4x^2+7)/7)|-(ln(7))/2))+C