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Title: Integration by parts problems and solutions
Description: Integration by parts is a technique used in calculus to evaluate the integral of a product of two functions. The formula for integration by parts stems from the product rule for differentiation and allows you to transform certain types of integrals into simpler ones.

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Integration by Parts (Exam Problems with Solutions):
1
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Solution: Let u = x and dv = e^x dx
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Applying integration by parts:
∫xe^x dx = xe^x - ∫e^x dx = xe^x - e^x + C
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Question: Find ∫x sin(x) dx
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Then, du = dx and v = -cos(x)
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3
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Solution: Let u = ln(x) and dv = dx
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Applying integration by parts:
∫ln(x) dx = x ln(x) - ∫(1/x) * x dx = x ln(x) - x + C
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Question: Evaluate ∫x^2 cos(x) dx
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Then, du = 2x dx and v = sin(x)
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5
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Solution: Let u = e^x and dv = sin(x) dx
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Applying integration by parts:
∫e^x sin(x) dx = -e^x cos(x) - ∫(-cos(x)) * e^x dx
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Question: Calculate ∫x^3 ln(x) dx
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Then, du = (1/x) dx and v = (1/4) x^4
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7
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Solution: Let u = x^2 and dv = e^x dx
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Applying integration by parts:
∫x^2 e^x dx = x^2 e^x - ∫2x e^x dx
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Question: Find ∫ln(x) / x dx
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Then, du = (1/x) dx and v = ln(x)
...

9
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Solution: Let u = x^4 and dv = sin(x) dx
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Applying integration by parts:
∫x^4 sin(x) dx = -x^4 cos(x) - ∫-4x^3 cos(x) dx
...
Question: Evaluate ∫x^3 e^x dx
Solution: Let u = x^3 and dv = e^x dx
...


11
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Solution: Let u = ln(x) and dv = cos(x) dx
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Applying integration by parts:
∫cos(x) ln(x) dx = ln(x) sin(x) - ∫(1/x) sin(x) dx
...
Question: Calculate ∫x^2 ln(x) dx
...
Then, du = (1/x) dx and v = (1/3) x^3
...

13
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Solution: Let u = x^3 and dv = cos(x) dx
...

Applying integration by parts:
∫x^3 cos(x) dx = x^3 sin(x) - ∫3x^2 sin(x) dx
Title: Integration by parts problems and solutions
Description: Integration by parts is a technique used in calculus to evaluate the integral of a product of two functions. The formula for integration by parts stems from the product rule for differentiation and allows you to transform certain types of integrals into simpler ones.