What is the integral of sqrt(1+x^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sqrt(1+x^2)

integral

1 + x^{2}

\frac{1}{2} x 1 + x^{2} + \frac{1}{2} ln x + 1 + x^{2} + C

Solution steps

\int 1 + x^{2} d x

Apply Trigonometric Substitution

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

Apply Integral Reduction

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

Substitute back

u = arctan (x)

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

Simplify

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

= \frac{1}{2} x 1 + x^{2} + \frac{1}{2} ln x + 1 + x^{2}

Add a constant to the solution

= \frac{1}{2} x 1 + x^{2} + \frac{1}{2} ln x + 1 + x^{2} + C

\int \frac{3}{sin ^{2} ( x )} d x

\int t (1.2)^{t} d t

\int e^{l n (s i n (x))} d x

\int \frac{1}{x \cdot ( x + 1 )} d x

\int (\frac{y}{5} + \frac{2}{y}) d y

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