-d x-9 x2-24 x-65 is equal toA- 1-7tan -13 x-4-27-CB- 1-21tan -13 x-4-2 x-7-CC- 1-4tan -12 x-7-1-21log3 x-4-7-C

The correct option is B 121tan^ 13x+47+C∫dx9x^2+24x+65 =∫dx(3x)^2+2×3x×4+4^2+49=∫dx(3x+4)^2+7^2 Let 3x+4=t ⇒ 3 dx=dt =13∫dtt^2+7^2=121tan^ 1t7+C =121tan^ 13x+ ...

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Byju's Answer

Standard XII

Mathematics

Functions

∫d x/9 x2+24 ...

Question

$\int \frac{d x}{9 x^{2} + 24 x + 65}$ is equal to

\frac{1}{7} {tan}^{- 1} \frac{3 x + 4}{7} + C

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\frac{1}{21} {tan}^{- 1} \frac{3 x + 4}{7} + C

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\frac{1}{4} {tan}^{- 1} \frac{2 x + 7}{4} + C

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\frac{1}{21} log \frac{3 x + 4}{7} + C

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Solution

The correct option is B $\frac{1}{21} {tan}^{- 1} \frac{3 x + 4}{7} + C$
$\int \frac{d x}{9 x^{2} + 24 x + 65} = \int \frac{d x}{(3 x)^{2} + 2 \times 3 x \times 4 + 4^{2} + 49} = \int \frac{d x}{(3 x + 4)^{2} + 7^{2}}$
Let $3 x + 4 = t \Rightarrow 3 d x = d t$
$= \frac{1}{3} \int \frac{d t}{t^{2} + 7^{2}} = \frac{1}{21} {tan}^{- 1} \frac{t}{7} + C = \frac{1}{21} {tan}^{- 1} \frac{3 x + 4}{7} + C$

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∫dx9x2+24x+65 is equal to

The correct option is B 121tan−13x+47+C∫dx9x2+24x+65=∫dx(3x)2+2×3x×4+42+49=∫dx(3x+4)2+72 Let 3x+4=t ⇒3 dx=dt =13∫dtt2+72=121tan−1t7+C=121tan−13x+47+C

$\int \frac{d x}{9 x^{2} + 24 x + 65}$ is equal to