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Calculus- Early Transcendentals, 2021a

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6.3. Indefinite Integrals 245<br />

Solution. The answer is:<br />

∫<br />

f (x)=<br />

f ′ (x)dx =<br />

∫ (x 4 + 2x − 8sinx ) dx<br />

=<br />

∫<br />

∫<br />

x 4 dx+ 2<br />

∫<br />

xdx− 8<br />

sinxdx<br />

where C is a constant.<br />

= x5<br />

5 + x2 + 8cosx +C,<br />

♣<br />

Example 6.23: Indefinite Integral<br />

∫<br />

Find the general indefinite integral of<br />

3x 2 dx.<br />

Solution.<br />

∫<br />

∫<br />

3x 2 dx = 3<br />

x 2 dx<br />

= 3 x3<br />

3 +C<br />

= x 3 +C<br />

♣<br />

Example 6.24: Indefinite Integral<br />

∫<br />

Find the general indefinite integral of<br />

2<br />

√ x<br />

dx.<br />

Solution. ∫ 2√x<br />

dx = 2∫<br />

x − 1 2 dx<br />

= 2 x− 1 2 +1<br />

− 1 2 + 1 +C<br />

= 4 √ x +C<br />

♣<br />

Example 6.25: Indefinite Integral<br />

∫ ( )<br />

1<br />

Find the general indefinite integral of<br />

x + e7x + x π + 7 dx.

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