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Math · Calculus I · Worked example

Handle a constant factor in u-substitution

Find ∫x²√(x³ + 1) dx.

∫x2x3+1d⁢x

Choose u

The expression under the root is x³ + 1, and its derivative 3x² matches the factor x² up to the constant 3. Let u = x³ + 1.

u=x3+1

Solve for the leftover factor

du = 3x² dx, so x² dx = du/3.

x2d⁢x=d⁢u3

Rewrite and integrate

The integral becomes (1/3)∫u^(1/2) du. Raise the power to 3/2 and divide by 3/2: (1/3)·(2/3)u^(3/2) = (2/9)u^(3/2).

13∫u1⁢/2d⁢u=29u3⁢/2+C

Substitute back

Replace u with x³ + 1.

29(x3+1)3⁢/2+C

Result

∫x²√(x³ + 1) dx = (2/9)(x³ + 1)^(3/2) + C.

Your turn

Find ∫x/(x² + 1) dx.

Show the answer and explanation

½ ln(x² + 1) + C.

Let u = x² + 1, so x dx = du/2. Then ∫(1/u)(du/2) = ½ ln|u| + C. Since x² + 1 is always positive, the absolute value can be dropped.

F⁢(x)=12ln(x2+1)F′(x)=xx2+1

Keep exploring

Open the rows in Math and change 2/9 to 2/3, the answer you get by forgetting the 1/3 from du. The box marks it as not an antiderivative.

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