What is the integral from 1 to 4 of (e^{3x}-1/x) ?

The integral from 1 to 4 of (e^{3x}-1/x) is (e^{12}-e^3)/3-2ln(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 4 of (e^{3x}-1/x)

\int_{1}^{4} (e^{3 x} - \frac{1}{x}) d x

\frac{e ^{12} - e ^{3}}{3} - 2 ln (2)

Decimal Notation

54243.51566 \dots

Solution steps

\int_{1}^{4} e^{3 x} - \frac{1}{x} d x

Apply the Sum Rule:

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

= \int_{1}^{4} e^{3 x} d x - \int_{1}^{4} \frac{1}{x} d x

\int_{1}^{4} e^{3 x} d x = \frac{e ^{12} - e ^{3}}{3}

\int_{1}^{4} \frac{1}{x} d x = 2 ln (2)

= \frac{e ^{12} - e ^{3}}{3} - 2 ln (2)

\int_{0}^{1} x 7^{x} d x

\int_{0}^{\frac{3 π}{2}} 2 + 2 sin (x) d x

\int_{0}^{5} \frac{1}{x ^{1.8}} d x

\int_{0}^{0.1} x e^{0.1 x} d x

\int_{0}^{1} x^{3} (1 + x)^{2} d x

What is the integral from 1 to 4 of (e^{3x}-1/x) ?
The integral from 1 to 4 of (e^{3x}-1/x) is (e^{12}-e^3)/3-2ln(2)