Integration by Parts Solver

Integration by Parts is a technique used in calculus to find the integral of a product of functions. It's essentially the reverse of the product rule for differentiation.

If you have an integral that can be written in the form $\int u(x) v'(x) dx$, Integration by Parts allows you to transform it into a potentially simpler integral.

The formula for Integration by Parts is:

$$ \int u \, dv = uv - \int v \, du $$

Where:

• $u$ is a function $u(x)$.
• $dv$ is another function $v'(x)dx$.
• $du$ is the derivative of $u$ with respect to $x$, times $dx$ (i.e., $u'(x)dx$).
• $v$ is the integral of $dv$ (i.e., $\int v'(x)dx$).

The key is to choose $u$ and $dv$ strategically so that the new integral, $\int v \, du$, is easier to solve than the original one.

The Integration by Parts formula can be derived from the product rule for differentiation.

The product rule states that for two differentiable functions $u(x)$ and $v(x)$:

$$ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} $$

We can write this in terms of differentials as:

$$ d(uv) = u \, dv + v \, du $$

Now, integrate both sides:

$$ \int d(uv) = \int u \, dv + \int v \, du $$

Since $\int d(uv) = uv$ (the integral of a differential is the function itself, up to a constant of integration which can be absorbed later), we have:

$$ uv = \int u \, dv + \int v \, du $$

Rearranging this equation to solve for $\int u \, dv$ gives us the Integration by Parts formula:

$$ \int u \, dv = uv - \int v \, du $$

The success of Integration by Parts often depends on choosing $u$ and $dv$ wisely. A common mnemonic to help with this choice is LIATE (or ILATE, DETAIL):

• Logarithmic functions (e.g., $\ln x, \log_b x$)
• Inverse trigonometric functions (e.g., $\arcsin x, \arctan x$)
• Algebraic functions (e.g., $x^2, 3x, \sqrt{x}$)
• Trigonometric functions (e.g., $\sin x, \cos x, \tan x$)
• Exponential functions (e.g., $e^x, 2^x$)

When choosing $u$, try to pick a function that appears earlier in the LIATE list. The function chosen for $u$ should simplify (or at least not get more complicated) when differentiated. The remaining part of the integrand, along with $dx$, becomes $dv$. The part chosen for $dv$ must be something you can integrate to find $v$.

For example, in $\int x \ln x \, dx$:

• $\ln x$ is Logarithmic (L).
• $x$ is Algebraic (A).
• Since L comes before A in LIATE, choose $u = \ln x$. This means $dv = x \, dx$.

Differentiating $u = \ln x$ gives $du = \frac{1}{x} dx$, which is simpler. Integrating $dv = x \, dx$ gives $v = \frac{x^2}{2}$, which is manageable.

The goal is for the new integral $\int v \, du$ to be easier to solve.

Example 1: $\int x e^x \, dx$

Using LIATE: A (Algebraic: $x$) comes before E (Exponential: $e^x$).

• Let $u = x \implies du = dx$
• Let $dv = e^x \, dx \implies v = \int e^x \, dx = e^x$

Applying the formula $\int u \, dv = uv - \int v \, du$:

$\int x e^x \, dx = x e^x - \int e^x \, dx$

$= x e^x - e^x + C$

$= e^x(x-1) + C$

Example 2: $\int \ln x \, dx$

This might look like it's not a product, but you can write it as $\int (\ln x) \cdot 1 \, dx$.

Using LIATE: L (Logarithmic: $\ln x$) comes before A (Algebraic: $1$).

• Let $u = \ln x \implies du = \frac{1}{x} dx$
• Let $dv = 1 \, dx \implies v = \int 1 \, dx = x$

Applying the formula:

$\int \ln x \, dx = (\ln x)(x) - \int x \left(\frac{1}{x}\right) dx$

$= x \ln x - \int 1 \, dx$

$= x \ln x - x + C$