What is the integral of 2^{0.1x^2} ?

The integral of 2^{0.1x^2} is 3.36614…\erfi(sqrt(ln(1.07177…))x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2^{0.1x^2}

\int 2^{0.1 x^{2}} d x

3.36614 \dots erfi (ln (1.07177 \dots) x) + C

Solution steps

\int 2^{0.1 x^{2}} d x

Simplify

2^{0.1 x^{2}} : 1.07177 \dots^{x^{2}}

= \int 1.07177 \dots^{x^{2}} d x

Use the nonelementary integral:

\int 1.07177 \dots^{x^{2}} d x = \frac{π erfi ( x ln ( 1.07177 \dots ) )}{2 ln ( 1.07177 \dots )}

= \frac{π erfi ( x ln ( 1.07177 \dots ) )}{2 ln ( 1.07177 \dots )}

Simplify

= 3.36614 \dots erfi (ln (1.07177 \dots) x)

Add a constant to the solution

= 3.36614 \dots erfi (ln (1.07177 \dots) x) + C

\int (cos (x))^{7} d x

\int \frac{sin ( 1 - x )}{1 - x} d x

\int y^{2} e^{s i n (2 x)}

\int 0.0003 x^{2} - 0.12 x + 20 d x

\int (3 x) d x

What is the integral of 2^{0.1x^2} ?
The integral of 2^{0.1x^2} is 3.36614…\erfi(sqrt(ln(1.07177…))x)+C