Integration: Learning Objective 6

Use integration by parts. A good choice is to let $u$ be the algebraic part ($x$) and $dv$ be the trigonometric part ($\sin(x)dx$).

\( -x\cos(x) + \sin(x) + C \)

We use the formula for integration by parts: \( \int u \, dv = uv - \int v \, du \).

Let $u = x$ and $dv = \sin(x)dx$.
Then we find $du = dx$ and $v = \int \sin(x)dx = -\cos(x)$.

Plugging these into the formula: \[ \int x\sin(x)\, dx = (x)(-\cos(x)) - \int (-\cos(x)) \, dx \] \[ = -x\cos(x) + \int \cos(x) \, dx \] \[ = -x\cos(x) + \sin(x) + C \]

Let $u=x$ and $dv = \cos(5x)dx$. Be careful when finding $v = \int \cos(5x)dx$.

\( \frac{x}{5}\sin(5x) + \frac{1}{25}\cos(5x) + C \)

Let $u = x$ and $dv = \cos(5x)dx$.
Then $du = dx$ and $v = \int \cos(5x)dx = \frac{1}{5}\sin(5x)$.

Using the integration by parts formula: \[ \int x\cos(5x)\, dx = (x)\left(\frac{1}{5}\sin(5x)\right) - \int \left(\frac{1}{5}\sin(5x)\right) \, dx \] \[ = \frac{x}{5}\sin(5x) - \frac{1}{5}\int \sin(5x) \, dx \] \[ = \frac{x}{5}\sin(5x) - \frac{1}{5}\left(-\frac{1}{5}\cos(5x)\right) + C \] \[ = \frac{x}{5}\sin(5x) + \frac{1}{25}\cos(5x) + C \]

This is a classic integration by parts problem. Let $u=\ln(x)$ and $dv=dx$. This choice works because the derivative of $\ln(x)$ is simpler.

\( x\ln(x) - x + C \)

Let $u = \ln(x)$ and $dv = dx$.
Then $du = \frac{1}{x}dx$ and $v = \int dx = x$.

Using the formula: \[ \int \ln(x)\, dx = (\ln(x))(x) - \int x \left(\frac{1}{x}dx\right) \] \[ = x\ln(x) - \int 1 \, dx \] \[ = x\ln(x) - x + C \]

Let $u = x$ and $dv = e^{-x}dx$.

\( -xe^{-x} - e^{-x} + C \) or \( -e^{-x}(x+1) + C \)

Let $u = x$ and $dv = e^{-x}dx$.
Then $du = dx$ and $v = \int e^{-x}dx = -e^{-x}$.

Using the formula: \[ \int xe^{-x}\, dx = (x)(-e^{-x}) - \int (-e^{-x}) \, dx \] \[ = -xe^{-x} + \int e^{-x} \, dx \] \[ = -xe^{-x} - e^{-x} + C \]

Choose $u = \ln(x)$ because its derivative is simpler. This leaves $dv = \frac{1}{x^2}dx$.

\( -\frac{\ln(x)}{x} - \frac{1}{x} + C \) or \( -\frac{1}{x}(\ln(x)+1) + C \)

Let $u = \ln(x)$ and $dv = \frac{1}{x^2}dx = x^{-2}dx$.
Then $du = \frac{1}{x}dx$ and $v = \int x^{-2}dx = -x^{-1} = -\frac{1}{x}$.

Using the formula: \[ \int \frac{\ln(x)}{x^2}\, dx = (\ln x)\left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right)\left(\frac{1}{x}dx\right) \] \[ = -\frac{\ln(x)}{x} + \int \frac{1}{x^2} \, dx \] \[ = -\frac{\ln(x)}{x} - \frac{1}{x} + C \]

This problem requires applying integration by parts twice. In the first step, let $u=x^2$. You will be left with another integral that requires integration by parts.

\( x^2e^x - 2xe^x + 2e^x + C \)

First Application of Integration by Parts:
Let $u = x^2$ and $dv = e^xdx$.
Then $du = 2x\,dx$ and $v = e^x$.
\[ \int x^2e^x\, dx = x^2e^x - \int e^x (2x\,dx) = x^2e^x - 2\int xe^x \, dx \]
Second Application of Integration by Parts:
We need to evaluate $\int xe^x \, dx$. Let $u = x$ and $dv = e^x dx$.
Then $du = dx$ and $v=e^x$.
\[ \int xe^x\, dx = xe^x - \int e^x \, dx = xe^x - e^x \]
Combine the Results:
Substitute the second result back into the first: \[ \int x^2e^x\, dx = x^2e^x - 2(xe^x - e^x) + C \] \[ = x^2e^x - 2xe^x + 2e^x + C \]

This problem is best solved by first using a substitution, followed by integration by parts. Let $u = -t^2$.

\( -\frac{1}{2} e^{-t^2}(t^2 + 1) + C \)

Step 1: Substitution
Let $u = -t^2$. Then $t^2=-u$ and $du = -2t\,dt$, so $t\,dt = -\frac{1}{2}du$.
Rewrite the integral by splitting $t^3$ into $t^2 \cdot t$: \[ \int t^3e^{-t^2} \, dt = \int t^2 e^{-t^2} (t \, dt) = \int (-u)e^u \left(-\frac{1}{2}du\right) = \frac{1}{2}\int u e^u \, du \]
Step 2: Integration by Parts
Now we evaluate $\int ue^u \, du$. Let $u_p=u$ and $dv_p = e^u du$.
Then $du_p = du$ and $v_p=e^u$.
\[ \int ue^u \, du = ue^u - \int e^u \, du = ue^u - e^u + C_1 \]
Step 3: Combine and Substitute Back
\[ \frac{1}{2}\int u e^u \, du = \frac{1}{2}(ue^u - e^u) + C = \frac{1}{2}e^u(u-1) + C \] Now substitute back $u=-t^2$: \[ = \frac{1}{2}e^{-t^2}(-t^2-1) + C = -\frac{1}{2}e^{-t^2}(t^2+1) + C \]

Use integration by parts with $u = \ln|ax+b|$ and $dv=dx$. The resulting integral will require another substitution.

\( \frac{ax+b}{a}\ln|ax+b| - x + C \) or \( (x + \frac{b}{a}) \ln|ax+b| - x + C \)

Let $u = \ln|ax+b|$ and $dv = dx$.
Then $du = \frac{a}{ax+b}dx$ and $v=x$.

\[ \int \ln|ax+b|\, dx = x\ln|ax+b| - \int x\left(\frac{a}{ax+b}\right) \, dx = x\ln|ax+b| - \int \frac{ax}{ax+b} \, dx \] To solve the remaining integral, we add and subtract $b$ in the numerator: \[ \int \frac{ax}{ax+b} \, dx = \int \frac{ax+b-b}{ax+b} \, dx = \int \left(\frac{ax+b}{ax+b} - \frac{b}{ax+b}\right) \, dx \] \[ = \int \left(1 - \frac{b}{ax+b}\right) \, dx = x - b \int \frac{1}{ax+b} \, dx = x - b\left(\frac{1}{a}\ln|ax+b|\right) \] Substitute this back into the original expression: \[ \int \ln|ax+b|\, dx = x\ln|ax+b| - \left(x - \frac{b}{a}\ln|ax+b|\right) + C \] \[ = x\ln|ax+b| - x + \frac{b}{a}\ln|ax+b| + C = \left(x+\frac{b}{a}\right)\ln|ax+b| - x + C \]

Integrate both sides of the equation for the prodcut rule with respect to $x$ and then rearrange the terms.

The proof starts with the Product Rule, integrates both sides, and algebraically rearranges the resulting equation to isolate one of the integrals, yielding the IBP formula.

1. Start with the Product Rule: For differentiable functions $f(x)$ and $g(x)$, we have: \[ \frac{d}{dx}[f(x)g(x)] = f(x)g'(x) + g(x)f'(x) \] 2. Integrate both sides with respect to x: \[ \int \frac{d}{dx}[f(x)g(x)] \, dx = \int (f(x)g'(x) + g(x)f'(x)) \, dx \] The integral of a derivative gives back the original function, so the left side simplifies: \[ f(x)g(x) = \int f(x)g'(x) \, dx + \int g(x)f'(x) \, dx \] 3. Rearrange the equation: Isolate one of the integrals, for example, $\int f(x)g'(x) \, dx$: \[ \int f(x)g'(x) \, dx = f(x)g(x) - \int g(x)f'(x) \, dx \] 4. Use differential notation: Let $u = f(x)$ and $v = g(x)$. Then $du = f'(x)dx$ and $dv = g'(x)dx$. Substituting these into the equation gives the standard form of the Integration by Parts formula: \[ \int u \, dv = uv - \int v \, du \]