What is the integral of+2x^2sec^2(x^3+1) ?

The integral of+2x^2sec^2(x^3+1) is 2/3 tan(x^3+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of+2x^2sec^2(x^3+1)

\int + 2 x^{2} sec^{2} (x^{3} + 1) d x

\frac{2}{3} tan (x^{3} + 1) + C

Solution steps

\int 2 x^{2} sec^{2} (x^{3} + 1) d x

Take the constant out:

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

= 2 \cdot \int x^{2} sec^{2} (x^{3} + 1) d x

Apply u-substitution

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

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

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= 2 \cdot \frac{1}{3} tan (u)

Substitute back

u = x^{3} + 1

= 2 \cdot \frac{1}{3} tan (x^{3} + 1)

Simplify

2 \cdot \frac{1}{3} tan (x^{3} + 1) : \frac{2}{3} tan (x^{3} + 1)

= \frac{2}{3} tan (x^{3} + 1)

Add a constant to the solution

= \frac{2}{3} tan (x^{3} + 1) + C

\int x 1 + 2 x d x

\int_{0}^{3} \frac{1}{5 x + 1} d x

\int_{0}^{5} (5 x - 1) d x

\int \frac{e ^{x} - e ^{- x}}{e ^{x} - e ^{- x}} d x

y = ∣ x ∣

What is the integral of+2x^2sec^2(x^3+1) ?
The integral of+2x^2sec^2(x^3+1) is 2/3 tan(x^3+1)+C