Home > Calculus > Definite and indefinite integrals

What is #int ln(x)^2+xdx#?

Answer 1

Elijah Abram

#x\ln (x^2)-2x+\frac{x^2}{2}+C#

#\int \ln (x^2)+xdx#

Applying sum rule: =#\int f(x)\pm g\(x)dx=\int f(x)dx\pm \int g(x)dx#

=#\int \ln (x^2)dx+\int \xdx#.... (i)

=#\int \ln (x^2)dx#

Applying integration by parts:

=#\int \uv'=uv-\int u'v#

=#u=\ln (x^2),\u'=\frac{2}{x},v'=1,v=x#

=#\ln (x^2)x-\int \frac{2}{x}xd#

=#x\ln (x^2)-\int 2dx# (#int 2dx#= 2x)

=#x\ln (x^2)-2x#

Also, =#\int xdx=\frac{x^2}{2}# (Applying power rule, #\int x^adx=\frac{x^{a+1}}{a+1}#)

Now substituting the value in equation (i),

=#x\ln (x^2)-2x+\frac{x^2}{2}#

=#x\ln (x^2)-2x+\frac{x^2}{2}# + C

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Aurora Alvarado

The integral of ln(x)^2 + x dx is not expressible in terms of elementary functions. It is a non-elementary integral and cannot be evaluated using basic functions like polynomials, exponentials, trigonometric functions, and their inverses.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.